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https://github.com/dimitrisjim/leetcode_solutions
Solutions to leetcode problems in Python, Rust, C and JavaScript.
https://github.com/dimitrisjim/leetcode_solutions
c javascript leetcode python rust
Last synced: 16 days ago
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Solutions to leetcode problems in Python, Rust, C and JavaScript.
- Host: GitHub
- URL: https://github.com/dimitrisjim/leetcode_solutions
- Owner: DimitrisJim
- License: unlicense
- Created: 2020-10-06T23:27:22.000Z (about 4 years ago)
- Default Branch: main
- Last Pushed: 2022-11-02T11:55:26.000Z (about 2 years ago)
- Last Synced: 2024-11-25T14:48:24.711Z (about 1 month ago)
- Topics: c, javascript, leetcode, python, rust
- Language: C
- Homepage:
- Size: 1.62 MB
- Stars: 3
- Watchers: 3
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# LeetCode problems.
Some small notes:
- I usually use the best execution/memory numbers I can achieve.
Sometimes this makes me write code which is what you would call
premature optimization (especially in Python.)
- Rust seems to optimize very well and uniformly, leading
to `100%` execution/memory consistently. (Update: Low number of submissions
also has an effect.
- Search for `TODO: Improve` for places where I think I've messed up slightly
(also see Revisit.md).
- Problems that are new have skewed timings because of the initial low
number of submissions.
- I might skip some `C` problems when the time needed to create the structures far outweights the benefit.
Most times it is relatively straight-forward after a solution has been found.
Additionally:
- Python and C also have a Concurrency folder where attempts at the concurrency problems are
added. I will not offer any sort of guarantees on those.
## [1. Two Sum.][1]
Similar to other Two Sum problems. We again use a cache in the form of a mapping
that keeps track of the appropriate index to use when we finally find the match.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 96.96 | 0 - 100.00 | 72 - 94.33 | 28 - 99.99 |
| Mem Usage (MB-%)| 6 - 68.00 | 2.3 - 68.01 | 38.4 - 91.28 | 14.4 - 47.87 |
## [2. Add Two Numbers.][2]
Traverse each list and build the counts as we go. Reuse one of the lists instead of
allocating new nodes all the time. All in all, this has `O(N)` worse case runtime complexity
where `N` is the size of the largest linked list.
Rust: Not sure how exactly I should do this :(.
TODO: Revisit.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 95.93 | - | 120 - 97.43 | 56 - 98.88 |
| Mem Usage (MB-%)| 7.2 - 96.67 | - | 44 - 72.19 | 14.2 - 90.81 |
## [7. Reverse Integer.][7]
We can work with arrays of the digits in order to reverse the numbers. Care needs to be
taken specifically for largest negative `i32 (0x80000000)` and for inputs that contain
`10` digits in total (which, when reversed, might not fit in 32 bits).
To combat first case, we can just bail early if `x == 0x80000000`. To combat the second case,
we check the digits of the result against the maximum value allowed. If, at any point, we discover
input the would result in a number too large to represent, we bail.
Overall, this is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 88 - 94.61 | 20 - 99.44 |
| Mem Usage (MB-%)| 6 - 14.40 | 1.9 - 85.78 | 40.8 - 14.78 | 14.4 - 11.58 |
## [9. Palindrome Number.][9]
Common string version, transform to string and then go through from both ends
trying to find a mismatch. `O(num_of_digits)` basically.
Without transforming to string: You could do something similar by divmoding to
find the first digits and divmoding again to find last. At least, that's what
I'm thinking now. TODO: look this up.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 92.79 | 0 - 100.00 | 196 - 96.90 | 48 - 94.22 |
| Mem Usage (MB-%)| 6.2 - 10.13 | 2.2 - 31.71 | 49.1 - 36.11 | 14.2 - 49.05 |
## [13. Roman to Integer.][13]
Keep a set of the special cases and go through each character in `s`. If the character
we've currently looking at and its next one are in the special cases, handle the case
and increase our counter to then look at the next character two positions over. Else,
just add the character seen.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 89.64 | 0 - 100.00 | 152 - 95.49 | 36 - 97.76 |
| Mem Usage (MB-%)| 5.7 - 98.67 | 2 - 92.20 | 44.4 - 92.17 | 14 - 94.68 |
## [14. Longest common prefix.][14]
First string is initialized as the prefix. We iterate through the rest of the strings
and continuously reduce this prefix. Process stops when all strings have been seen or
if the prefix becomes empty at some point during the iteration.
Space complexity it `O(1)`, time complexity is `O(NM)` where `N` is the number of
strings and `M` the length of the strings.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 97.82 | 28 - 94.45 |
| Mem Usage (MB-%)| 5.8 - 96.22 | 2.1 - 75.69 | 39.7 - 62.36 | 14 - 99.12 |
## [21. Merge two sorted lists.][21]
For `Rust`, go through both linked lists and build vector of values, then, build
resulting vector.
For others, go through nodes of linked list and incrementally build resulting linked
list.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 4 - 100.00 | 84 - 92.62 | 32 - 91.62 |
| Mem Usage (MB-%)| 6.3 - 47.94 | 2 - 83.19 | 40.1 - 91.86 | 14.1 - 83.79 |
## [24. Swap nodes in pairs.][24]
Relatively straight-forward though the swapping is easy to get wrong. Keep track
of previous in line (to point to new next), the current pairs (cur, next) we are
swapping and perform the swap.
This touches each node once so `O(N)` time with `O(1)` space.
Rust: I probably need to read Too Many Linked Lists, again.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | - | 72 - 92.46 | 20 - 99.11 |
| Mem Usage (MB-%)| 5.8 - 60.54 | - | 38.7 - 71.53 | 14.3 - 48.70 |
## [27. Remove Element.][27]
Use a swap-remove. As found in Rusts `Vec.swap_remove`. Basically, replace the
value you want to remove with the value at the end of the list/array. This way,
`O(1)` removes can be achieved. Overall complexity is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 98.77 | 28 - 90.92 |
| Mem Usage (MB-%)| 6.1 - 39.77 | 2.1 - 67.18 | 38.6 - 60.38 | 14.3 - 46.75 |
## [28. Implement Strstr.][28]
After handling special edge cases (`needle == ""` and `len(needle) > len(haystack)`) we
go through the haystack and compare sub-slices. Need to break out early if a subslice doesn't
match. Worse case is still `O(nm)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 99.08 | 24 - 95.52 |
| Mem Usage (MB-%)| 5.8 - 95.27 | 2.2 - 58.11 | 39.4 - 41.95 | 14.5 - 24.15 |
## [31. Next Permutation.][31]
This basically uses the algorithm described [here](https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order).
In short, find index `i` such that for all `j` > `i` `nums[j] > nums[j+1]`. Swap the value with the smallest value that's larger
than the value at index `i`. Reverse the contents of the array for `j > i`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 98.86 | 0 - 100.00 | 76 - 94.00 | 28 - 99.78 |
| Mem Usage (MB-%)| 6.1 - 78.86 | 2 - 56.25 | 40.5 - 28.99 | 14.3 - 52.14 |
## [34. Find first and last position of element in sorted array.][34]
Use binary searches to find start/end.
C: todo, too bored to implement binary search (again!)
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 6 - 91.00 | 75 - 99.81 |
| Mem Usage (MB-%)| - | 2.3 - 78.42 | 3.9 - 99.90 | 15.4 - 49.40 |
## [35. Search insert position.][35]
Again, binary searching for the right spot, i.e `O(logN)`.
Start points to position in list for which all elements with index smaller than start have a value
than target and all elements with index larger than it have a `value >= target`.
When finished, start will either point to target or the leftmost position in which to insert it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 98.11 | 40 - 97.25 |
| Mem Usage (MB-%)| 6 - 85.77 | 2 - 100.00| 38.7 - 72.22| 15.1 - 52.12 |
## [46. Permutations.][46]
This basically uses the algorithm described [here](https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order).
In short, find index `i` such that for all `j` > `i` `nums[j] > nums[j+1]`. Swap the value with the smallest value that's larger
than the value at index `i`. Reverse the contents of the array for `j > i`.
The only difference between this problem and `31.` is that here we stick it all in a loop and we sort the sequence before
we enter the loop (in order to start from 'smallest' permutation).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 95.45 | 0 - 100.00 | 80 - 96.09 | 32 - 97.03 |
| Mem Usage (MB-%)| 7.1 - 85.61 | 2.1 - 86.49 | 41.8 - 49.81 | 14.3 - 70.48 |
## [50. Pow(x, n).][50]
Uses binary exponentiation (exponentiation by squaring) to cut down on the number
of multiplications performed to `O(logn)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.48 | 24 - 96.07 |
| Mem Usage (MB-%)| 5.5 - 61.56 | 2 - 88.89 | 40.5 - 8.33 | 14 - 92.69 |
## [53. Maximum Subarray.][53]
Can keep track of largest running sum while going through the array. Initially, I used a table to keep track
of the values seen as the following code illustrates:
```python
def maxSubArray(self, nums: List[int]) -> int:
length, maximum = len(nums), nums[0]
table = [0] * len(nums)
table[0] = nums[0]
for i in range(1, length):
cur, prev = nums[i], table[i-1]
if (s := cur + prev) > cur:
table[i] = s
else:
table[i] = cur
if table[i] > maximum:
maximum = table[i]
return maximum
```
Of course, after closer examination, we can see we only need to know the previous value in order to keep track of the
running sum (we always access `table[i-1]` and `table[i]`, hinting that one variable whose value we update could do the
trick.)
Uses `O(N)` time and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 97.38 | 0 - 100.00 | 68 - 99.79 | 60 - 89.66 |
| Mem Usage (MB-%)| 6.2 - 97.47 | 2.1 - 97.32 | 38.9 - 87.32 | 14.9 - 78.41 |
## [58. Length of last word.][58]
If the length is easily (`O(1)`) available, walk the string backwards and return the second time we reach
a space. If length isn't available, traverse until the end continuously updating the count per word and return
the final value of count stored. All in all, `O(N)` runtime with `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 99.72 | 20 - 98.80 |
| Mem Usage (MB-%)| 5.8 - 55.26 | 2.1 - 78.18 | 38.5 - 58.54 | 14.4 - 6.90 |
## [61. Unique paths.][61]
DP with tabulation. Start with an initial value of `1` in 2d table position `[0][0]` (start) and then add
values from the left and top as the table is traversed (signifying moves made right and down).
Final answer is at last position (end.)
Space and time complexity is `O(mn)`.
Note: there's definitely a space optimization here. Only work with 2 rows at a time.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 98.41 | 20 - 99.30 |
| Mem Usage (MB-%)| 5.8 - 24.15 | 2 - 94.44 | 39.6 - 23.46 | 14.2 - 65.47 |
## [63. Unique paths II.][63]
DP with tabulation. Start with an initial value of `1` in 2d table position `[0][0]` (start) and then add
values from the left and top as the table is traversed (signifying moves made right and down). Make sure
to ignore positions with obstacles and to only add if we came from free positions.
Final answer is at last position (end.)
Space and time complexity is `O(mn)`.
Note: there's definitely a space optimization here. Only work with 2 rows at a time.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 98.42 | 36 - 94.99 |
| Mem Usage (MB-%)| 6 - 71.08 | 2.1 - 60.00 | 39.5 - 45.34 | 14.1 - 94.87 |
## [64. Summary Ranges.][64]
Keep a tracker around to track the sequence as it increases. If at any point, the number in
the array doesn't match the tracker, a new sequence should be added. Care is needed when
ending the loop to accound for two cases (ended on a single number or while going through a
range). Overall, `O(N)` time and space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 97.06 | 24 - 93.79 |
| Mem Usage (MB-%)| 5.7 - 73.33 | 2.1 - 54.55 | 38.2 - 92.51 | 14.4 - 16.10 |
## [66. Plus One][66]
Relatively straight-forward, if we encounter a `9`, we need to zero it and move
left to the next item in the array. If it isn't a nine, we increment and return.
Care is needed to detect when the array is basically full of `9`s in which case we
need to prepend a `1` before we return.
Runtime complexity is `O(N)`, we perform constant operations for each item in the
array. Space complexity is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 99.75 | 24 - 97.53 |
| Mem Usage (MB-%)| 5.8 - 99.54| 2 - 93.75 | 38.6 - 69.35 | 14.3 - 48.72 |
## [69. Sqrt(x).][69]
Use [babylonian method](https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method)
for finding square roots. Initial guess is `x / 2` and we converge by taking the arithmetic mean of
`guess, x / guess` until `guess * guess` is `<= x`.
I am not sure about the complexity, a quick search yields `O(log(log(n/m)))` where `n` is our input
and `m` is the error during the approximation. Since we don't mind large errors (require integer result)
`n/m` small (i.e nearing `n`).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 99.77 | 28 - 94.96 |
| Mem Usage (MB-%)| 5.5 - 90.98 | 2 - 91.18 | 40.2 - 21.04 | 14.3 - 6.07 |
## [70. Climbing stairs.][70]
Pretty much similar to finding the `n`th fibonacci number. `O(N)` time with `O(1)` space (we only need
to keep track of previous two values.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 96.42 | 24 - 92.99 |
| Mem Usage (MB-%)| 5.4 - 99.04 | 2 - 87.23 | 38.3 - 74.43 | 13.8 - 99.83 |
## [74. Search a 2d matrix.][74]
Perform a binary search on each of the rows. `O(nlogn)` runtime and `O(1)` space. Also solves
problem `240` but not so efficiently.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.05 | 32 - 98.77 |
| Mem Usage (MB-%)| 6.1 - 84.76 | 2.1 - 78.57 | 38.5 - 87.74 | 14.8 - 62.06 |
## [78. Subsets.][78]
Trick also utilized in Knuth 4A, view subsets as bitstrings of length `size(input)` with members
being denoted by a set bit for the appropriate position. That is, for input of the form `[1, 2, 3, 4]`
we'd have:
```
0000 0001 0010 0100 1000 0011 0101 1001 0110 1010 1100 0111 ...
```
where these denote the subsets:
```
{0}, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {3, 4}, {1, 2, 3} ...
```
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 99.13 | 0 - 100.00 | 72 - 98.13 | 28 - 92.84 |
| Mem Usage (MB-%)| 6.5 - 65.22 | 2 - 90.00 | 39.6 - 73.46 | 14.4 - 51.70 |
## [83. Remove duplicates from sorted list.][83]
Traverse and update the next pointers/references. `O(N)` time.
Rust: LinkedList chaos.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | - | 72 - 99.25 | 36 - 94.09 |
| Mem Usage (MB-%)| 6.4 - 93.20 | - | 40.6 - 44.91 | 14 - 94.35 |
## [88. Merge sorted array.][88]
Place elements from `nums2` at the end of `nums1` (replacing zero elements) and sort it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 98.93 | 28 - 96.98 |
| Mem Usage (MB-%)| 6.2 - 39.46 | 2 - 66.67 | 38.5 - 92.89 | 14.2 - 85.77 |
## [94. Binary Tree Inorder Traversal.][94]
Python + JS use iterative approach. C + Rust use recursive. With C, going iterative is
relatively trivial, for Rust, not so much (and I'm really not sure how you'd do it without
having TreeNode be clone'able [and even that sounds like a nasty approach]).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 93.61 | 12 - 95.37 |
| Mem Usage (MB-%)| 5.7 - 95.00 | 1.9 - 94.29 | 38.5 - 79.11 | 13.3 - 75.85 |
## [100. Same Tree][100]
C recursively solves it while the others use the iterative approach. In all cases
this really just involves traversing the same nodes and checking that their values
are the same and they both have the same numbers of children.
Care is needed to handle case where one tree has been exhausted while the other tree
still has nodes we haven't seen.
Complexity is `O(N)` where N is the number of nodes since we perform constant
operations for each node. Not sure about space complexity of the top of my head.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 92.99 | 24 - 95.46 |
| Mem Usage (MB-%)| 6 - 34.24 | 2 - 52.17 | 38.8 - 77.73 | 14.2 - 65.65 |
## [101. Symmetric Tree.][101]
Walk both subtrees in a symmetric way (when we walk left on one, we walk right on
the other and vise versa). This relationship can be captured either via a normal
stack or the call stack of a function.
Python, Rust, Javascript use a list/vector to order the traversal down both branches while
C is recursive. `O(N)` for either case (memory and runtime)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 84 - 90.33 | 28 - 92.94 |
| Mem Usage (MB-%)| 6.7 - 98.62 | 2 - 73.53 | 40.6 - 35.90 | 14.6 - 10.69 |
## [102. Binary Tree level order traversal.][102]
There's a similar problem to this one, I remember it. All in all, use a stack where we
keep nodes/level tuples. DFS traversal and use the level saved for each node as index in
the results vec/list that holds the result values. `O(N)` space/memory.
C: Too much hassle with 2D arrays but I hope I'll get back to it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 0 - 100.00 | 80 - 85.22 | 24 - 98.58 |
| Mem Usage (MB-%)| | 2 - 97.96 | 40.3 - 44.58 | 14.5 - 88.37 |
## [104. Maximum depth of Binary Tree][104]
Rust contains both iterative and recursive approaches. C/Javascript use
recursive, Python uses iterative.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 92.16 | 0 -100.00 | 80 - 94.64 | 36 - 89.43 |
| Mem Usage (MB-%)| 8.1 - 50.63 | 2.5 - 94.87 | 41 - 96.55 | 15.2 - 91.21 |
## [107. Binary Tree Level Order Traversal II,][107]
Breadth first traversal, use an additional value to mark the level which we use to
correctly place into subarrays. Return result reversed.
C Version: Not mentally ready to go through the allocation hell, yet.
Complexity: Time should be `O(N)`, not sure about space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| Nope. | 0 - 100.00 | 76 - 95.89 | 24 - 98.69 |
| Mem Usage (MB-%)| NOPE. | 2.1 - 88.89 | 40.3 - 25.94 | 14.7 - 49.77 |
## [108. Convert Sorted Array to BST.][108]
Need to balance resulting tree (i.e max difference of height between two children of
a given node to equal 1).
Since we have a sorted array, we know we can separate the array in a triplet consisting
of `(lower_than_mid, mid, higher_than_mid)`. Using this we can then use:
- `mid`: as our new node.
- `lower_than_mid`: as our left branch.
- `higher_than_mid`: as our right branch.
Recursively solving this is straight-forward.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 94.69 | 0 - 100.00 | 84 - 95.62 | 56 - 99.93 |
| Mem Usage (MB-%)| 18 - 59.59 | 2.9 - 71.43 | 41.7 - 93.78 | 16.6 - 22.27 |
## [110. Balanced Binary Tree.][110]
Need to go through tree and calculate differences between subtrees for every given node.
Doing this naively results in a lot of repetition of work (think fibonacci). So, a cache
can be utilized that holds the height for a given node.
Approaches here use recursion, C version is TODO for when I'm up for using uthash again. Got
Rust version to work but cache requires raw pointers as keys, I feel a better approach should
exists and I'm under the impression an iterative solution would be better. As such, that is
another TODO.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 4 - 100.00 | 92 - 77.70 | 40 - 97.82 |
| Mem Usage (MB-%)| - | 2.9 - 12.50 | 43 - 62.04 | 18.9 - 48.15 |
## [111. Minimum depth of binary tree.][111]
Typical depth first recursion. We can stop recursing when we reach a depth which is bigger
than a minimum we've already found, though. Worse case complexity is still `O(N)` as is the
memory complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 132 - 99.58 | 32 - 97.73 | 232 - 99.61 | 468 - 93.47 |
| Mem Usage (MB-%)| 75.4 - 99.16 | 11.4 - 100.00 | 75.4 - 17.90 | 49.3 - 72.86 |
## [112. Path Sum.][112]
Rust and Python have iterative solutions while C/Javascript have recursive. The idea
in both cases is the same, recurse while holding on to the current sum and, when we
reach a leaf, check if sums match.
`O(N)` both space and runtime complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 96.72 | 0 - 100.00 | 80 - 97.72 | 36 - 93.49 |
| Mem Usage (MB-%)| 8.4 - 15.54 | 2.5 - 92.31 | 42.2 - 31.94 | 16 - 24.23 |
## [113. Path Sum II.][113]
Rust and Python have iterative solutions while Javascript have recursive. The idea
in both cases is the same, recurse while holding on to the current sum, the level (depth) and
the path of nodes encountered. When we encounter a level with value `<=` then the length of the
list of nodes (path), we trim the path to bring the path to the correct position.
When we reach a leaf, check if sums match and, if they do, yield/store the path.
`O(N)` both space and runtime complexity.
TODO: `C` version isn't hard just requires manual management for the arrays holding paths, list
of paths which I've burned out doing over and over.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 88 - 95.02 | 36 - 97.12 |
| Mem Usage (MB-%)| - | 2.5 - 100.00 | 49.6 - 25.25 | 14.9 - 99.21 |
## [118. Pascal's triangle.][118]
Becomes easy once you look up on Pascal's Triangle and find the [formula for
finding a row](https://en.wikipedia.org/wiki/Pascal%27s_triangle#Calculating_a_row_or_diagonal_by_itself).
Not sure about complexity here. Since for each `n` we perform `1 + 2 + 3 + .... + n`
iterations, though, I believe it must be around `O(n^2)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 99.32 | 24 - 95.43 |
| Mem Usage (MB-%)| 5.9 - 95.86 | 2.3 - 6.67 | 38.6 - 21.82 | 14.3 - 59.05 |
## [116. Populating next right pointers in each node.][116]
The only issue is when you need to link two "cousins" (children of uncle nodes) together.
In order to do that, we just use `node->next` that has been filled in previous recursion/iteration
to go to the uncle and access his children. Care needs to be taken on what we link right has priority
on the left side while left has priority on the right side, i.e:
```
# link 3 to 4 but if 3 doesn't exist, link 9 to 4 and
# if 4 doesn't exist, link 3 or 9 to 5.
root
/ \
4 7
/ \ / \
9 3 4 5
```
Complexity is `O(N)` both runtime and space (counting implicit call stack).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 83.52 | N/A | 92 - 95.75 | 52 - 97.91 |
| Mem Usage (MB-%)| 8.8 - 97.80 | N/A | 45.2 - 72.90 | 15.5 - 99.07 |
## [119. Pascal's triangle II.][119]
Becomes easy once you look up on Pascal's Triangle and find the [formula for
finding a row](https://en.wikipedia.org/wiki/Pascal%27s_triangle#Calculating_a_row_or_diagonal_by_itself).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 99.83 | 24 - 96.72 |
| Mem Usage (MB-%)| 5.9 - 65.63 | 2.4 - 5.00 | 38.3 - 82.37 | 14.3 - 54.93 |
## [121. Best time to buy and sell stock.][121]
Continuously holds the minimum encountered thus far. When a
value is not smaller, we see if it exceeds the max_price as
has been set thus far. For example:
[1, 2, 3, 0, 20, 0, 30, 5, 6]
min_price is initialized to `prices[0]` first (by default, the minimum
encountered thus far). Our max_price during the iteration tracks the
following subtractions:
price == 1: => min_price = 1, max_price = 0 (dummy iteration, really)
price == 2: => min_price = 1, max_price = 1 (2 - 1)
price == 3: => min_price = 1, max_price = 2 (3 - 1)
....
price == 30 => min_price = 0, max_price = 30 (30 - 0). (Final answer)
Runs in O(N) with O(1) space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 96 - 100.00 | 8 - 97.35 | 92 - 93.75 | 952 - 83.52 |
| Mem Usage (MB-%)| 13 - 43.30 | 3.1 - 23.18 | 48.5 - 90.30 | 25 - 95.24 |
## [122. Best time to buy and sell stock II.][122]
Buy on local minimums and sell on local maximums. This can be done by going through
the prices in pairs and buying only if the price of the previous day was smaller
than that of the next day and only selling if the price of the previous day is larger
than that for the next.
A toggle for the action is used to differentiate the cases (buying/selling).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 97.15 | 0 - 100.00 | 76 - 92.04 | 52 - 96.93 |
| Mem Usage (MB-%)| 6.9 - 46.64 | 2.1 - 76.47 | 39.2 - 72.89 | 15.1 - 54.54 |
## [125. Valid Palindrome.][125]
Strip unwanted characters, transform to uppercase and check if the remaining characters form a palindrome.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 3 - 82.32 | 37 - 99.32 |
| Mem Usage (MB-%)| 6.3 - 73.04 | 2.2 - 91.42 | 3.3 - 45.58 | 15.3 - 24.50 |
## [133. Clone graph.][133]
Recurse on neighbors and gradually build the graph, need to keep track of what we've visited
in order to not recurse like crazy. Both time and space complexity should be `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 87.32 | N/A | 72 - 98.91 | 32 - 92.73 |
| Mem Usage (MB-%)| 6.9 - 76.06 | N/A | 39.7 - 98.91 | 14.7 - 26.45 |
## [136. Single Number][136]
Pretty well known xor trick. Based on the property that xor of two equal
numbers is zero.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 99.30 | 0 - 100.00 | 80 - 92.47 | 112 - 99.63 |
| Mem Usage (MB-%)| 7.1 - 99.80 | 2.1 - 81.55 | 39.9 - 98.51 | 16.7 - 29.16 |
## [141. Linked list cycle.][141]
You could use two pointers, one ahead of the other and keep iterating until they
meet (or a `None/NULL/null` is encountered). If doing destructive mutations on the
linked list was allowed, you could alter the next pointer of all nodes to a dummy
node and iterate until you meet that or a `None/NULL/null` (but I doubt that
would be a good idea in general).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 99.80 | N/A | 80 - 94.82 | 44 - 97.97 |
| Mem Usage (MB-%)| 7.7 - 95.40 | N/A | 41 - 72.35 | 17.5 - 91.05 |
## [144. Binary Tree Preorder Traversal.][144]
Iterative approach for Python/Rust/Javascript. C uses recursive in order to avoid
managing memory for stack needed in that case (which is implicitly provided in recursive
case).
All display `O(N)` runtime/memory complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 94.45 | 24 - 94.50 |
| Mem Usage (MB-%)| 6 - 52.38 | 2 - 55.56 | 38.7 - 62.16 | 14 - 91.16 |
## [145. Binary Tree Postorder Traversal.][145]
Iterative approach for Python/Javascript. C and Rust use recursive solutions. For C, the
changes in order to make it iterative are relatively trivial, for Rust, not sure how easily you could
make it work iteratively.
All display `O(N)` runtime/memory complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 98.51 | 24 - 94.52 |
| Mem Usage (MB-%)| 5.9 - 61.46 | 2 - 94.12 | 38.6 - 59.70 | 14.1 - 74.99 |
## [155. Min Stack.][155]
See each file for detailed comments. Basically, hold min around and update it if we
pop it off at any point.
All operations other than pop (which is `O(N)`) are done in `O(1)` or amortized `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 24 - 97.52 | 0 - 100.00 | 112 - 96.23 | 48 - 99.24 |
| Mem Usage (MB-%)| 12.6 - 93.39 | 5.5 - 96.15 | 45.4 - 71.54 | 17.9 - 66.92 |
## [167. Two sum II input array is sorted.][167]
Array is sorted, use two counters (start, end) and increase, reduce respectively
until we reach our target sum.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 99.02 | 0 - 100.00 | 72 - 97.24 | 56 - 95.51 |
| Mem Usage (MB-%)| 6.6 - 99.02 | 2.1 - 60.71 | 38.8 - 71.96 | 14.5 - 91.92 |
## [171. Excel Sheet column number][171]
To find the column number we can follow the following formula where `s` is the input
string and `n` is its length:
```
col = sum((charCodePoint - 64) * (pow(26, i)) for i in [0..n])
```
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 88 - 95.33 | 24 - 96.97 |
| Mem Usage (MB-%)| 5.6 - 69.08 | 2 - 100.00 | 39.8 - 84.74 | 14.2 - 66.98 |
## [190. Reverse Bits.][190]
Count how many leading zeros exist in a 32bit padded (with zeros from MSB) version of the input number. Then just
add the bits from the input number to the result. This is done by shifting the result left by one and adding `n & 1` to
it and then shifting `n` to the right. Finally, we shift right as many times as we counted for the leading zeroes.
Javascript: couldn't adapt same solution. TODO.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | - | 28 - 87.17 |
| Mem Usage (MB-%)| 5.3 - 75.25 | 2 - 71.43 | - | 14.3 - 8.01 |
## [191. Number of 1 bits.][191]
Not much point in adding things here. The most efficient implementation is discussed in the [wiki page](https://en.wikipedia.org/wiki/Hamming_weight#Efficient_implementation) which also discusses a good strategy to optimize by using a cache of the result for all 16
bit sequences (which we can use for 32bit numbers by viewing them as two 16bit sequences).
For all intents and purposes, the algorithm I would run (if the language doesn't offer an implementation (rust, python do)) would be the
common iterative `O(N)`:
int count = 0;
while (n) {
n &= n - 1;
count += 1;
}
return count;
## [199. Binary Tree Right Side View.][199]
Traverse tree rightwards assigning levels to each node encountered. By using a stack
we can always push the rightmost values first and register them as we encounter them.
For each new level, we always register the rightmost node present in that level.
Overall needs `O(N)` execution time (since we traverse the full tree) and `O(N)` worse
case memory.
C: Could easily do this if I didn't despise having to manually track dynamic memory.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 72 - 99.10 | 20 - 99.43 |
| Mem Usage (MB-%)| - | 2.1 - 66.67 | 40.3 - 63.83 | 14.4 - 19.91 |
## [201. Bitwise and of numbers range.][201]
First interesting property, took me some time to notice.
If there's a difference in 1 bit it means, in the range
`[left, right]` there exists a number with the binary repr
`100....000` (`right.bit_length() - 1` zeroes) that completely
zeroes out all numbers smaller than it. The result, regardless
the range, will be 0. As a consequence, numbers that don't have a bit
representation of equal length automatically result to zero.
Second interesting property, way more time to notice.
The result, when the number of bits in the numbers is equal, is
equal to the binary number who's prefix is equal to the equal
prefixes of left and right. The rest of the bits are zero.
An example might better illustrate this:
```
left = 2146738493, right = 2147483647
```
And, in binary:
```
left = '1111111111101001010000100111101'
right = '1111111111111111111111111111111'
```
The result will be:
```
'1111111111100000000000000000000' == 2146435072
```
I.e the common prefix of left and right, `'11111111111'` followed by zeroes `'00000000000000000000'`.
This makes sense in the same way the first observation does. In the range between `left` and `right`,
the different combinations of `0` and `1` completely cancel each other out (and, by definition, all
combinations will exist between `'11110..'` and `'11111..'`).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 95.59 | 0 - 100.00 | 160 - 100.00 | 36 - 99.83 |
| Mem Usage (MB-%)| 5.7 - 97.79 | 2 - 72.73 | 47 - 24.00 | 14.2 - 83.45 |
## [206. Reverse Linked List][206]
`C` contains both iterative and recursive. The rest use iterative.
Note: Needless to say, Rust timings seem a bit off! Check again if you see
this in the future, future me.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 81.03 | 280 - 100.00 | 80 - 84.82 | 32 - 85.73 |
| Mem Usage (MB-%)| 6.2 - 98.35 | 2.5 - 51.56 | 40 - 94.38 | 15.4 - 69.73 |
## [217. Contains Duplicate.][217]
Can use a set and check if we've seen an element before (or, build a full set and
compare its size against that of the size of the original array.)
(`C` and `uthash` again seem slow)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 24 - 41.52 | 0 - 100.00 | 76 - 98.14 | 108 - 96.99 |
| Mem Usage (MB-%)| 17 - 10.73 | 2.7 - 84.62 | 45 - 54.84 | 20.1 - 81.21 |
## [219. Contains Duplicates II.][219]
Keep track of the most recent index seen in a map. When we encounter a value already contained
in the map, check if the diff of the indices is smaller than `k`. If so, return `true`, else, update
the `index` to hold the newly seen index. `O(N)` both space and runtime complexity.
For C: uthash can easily be used. A list-as-a-map cannot, as far as I can see, values contained exceed
list size.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 76 - 99.60 | 88 - 91.05 |
| Mem Usage (MB-%)| - | 4.2 - 28.57 | 44.5 - 58.12 | 21.8 - 54.83 |
## [220. Contains Duplicate III.][220]
Sort the numbers array after transforming it into a list holding [index, value] pairs
(enumerate its values). After doing that, we can start from the beginning and check
pairs for which the difference in value is `<= t`, larger values aren't considered.
For each pair for which `v2 - v1 <= t`, we check if the diff of their indices is
` <= k`.
Worse case is `O(N ** 2)` but that manifests only for extremely large values of `t`.
Note: I'm under the impression that another way might exist. My lackluster memory timings
and not being able to crack 90% with Python hints at this.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 100.00 | 0 - 100.00 | 108 - 92.61 | 104 - 82.89 |
| Mem Usage (MB-%)| 8 - 7.69 | 2.4 - 33.33 | 45.6 - 6.82 | 18.1 - 11.80 |
## [225. Implement Stack Using Queues.][225]
Don't particularly like these problems (also see Queue using Stacks). I'll just
add the Python solution, not really planning on adding others.
Uses a single queue actually since I really couldn't see the point of using a
second queue instead of rotating, adjusting to use second queue is straight-forward.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| N/A | N/A | N/A | 24 - 94.57 |
| Mem Usage (MB-%)| N/A | N/A | N/A | 14.4 - 46.41 |
## [226. Invert Binary Tree][226]
Most (`C`, `Javascript` and `Rust` (because I hate myself, apparently))
use recursion to swap the nodes. Python uses an iterative approach by
utilizing a stack.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.23 | 20 - 99.04 |
| Mem Usage (MB-%)| 5.8 - 99.02 | 2 - 88.89 | 38.9 - 55.11 | 14.2 - 39.78 |
## [231. Power of two.][231]
`x = 2 ** n <=> n = log2(x)`. For non-powers of two `log2(x)` is not an integer. This boils down
to checking if `log2(x)` is an integer which can be done with `ceil(log2(x)) == log2(x)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 98.66 | 24 - 95.94 |
| Mem Usage (MB-%)| 5.5 - 56.84 | 2.5 - 6.25 | 40.2 - 13.07 | 14.2 - 40.52 |
## [232. Implement Queue using Stacks.][232]
Only uses stack equivalent operations. Two lists are created, one that holds values
pushed and one that holds values that will be popped. When time comes to pop/peek,
we check to see if the list holding values to be popped is empty.
If it's emtpy, we continuously push the values popped from the other list (thereby
restoring the FIFO principle).
If it isn't empty, we immediatelly just pop/peek from it.
`is_empty` is `O(1)`. The rest are amortized `O(1)` operations (`push` definitely is,
at least. I'm relatively certain `peek` and `pop` are too).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 99.36 | 20 - 98.63 |
| Mem Usage (MB-%)| 5.8 - 96.97 | 1.9 - 100.00 | 38.4 - 67.68 | 14.1 - 98.99 |
## [237. Delete node in a Linked List][237]
Swap contents of `node` with contents of `node.next`. Rust version not
available.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 95.10 | N/A | 80 - 94.74 | 20 - 100.00 |
| Mem Usage (MB-%)| 6.3 - 99.87 | N/A | 40.2 - 84.74 | 14.7 - 19.26 |
## [240. Search a 2-d matrix II.][240]
Start from bottom left corner. Eliminate rows by checking if `target < row[0]` and columns by
checking if `target > row[0]`. (All values on the right of `target` if `target < row[0]` must be
larger while if `target > row[0]` columns to the right will contain larger values.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 48 - 91.20 | 0 - 100.00 | 316 - 72.89 | 152 - 97.52 |
| Mem Usage (MB-%)| 9.1 - 92.00 | 2.5 - 78.95 | 41.7 - 81.98 | 20.4 - 97.73 |
## [242. Valid anagram.][242]
All except for C use a counter. No idea why `Rust` can't go lower than `4` though
I believe I could get it there with the same hack as in `C`.
Note: `C` case just seems hacky as hell, read comments (which do a pretty poor job
of explaining).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 4 - 61.40 | 84 - 95.92 | 28 - 99.36 |
| Mem Usage (MB-%)| 5.8 - 85.24 | 2.1 - 100.00 | 40.9 - 64.49 | 14.5 - 69.02 |
## [257. Binary Tree Paths,][257]
Recursively build path by holding (and passing) the partially constructed path
as an argument to the recursive function.
The path is added to the result once we reach a leaf.
Runtime complexity should be `O(N)` since we traverse each node. Space complexity
is related to the number of branches.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 99.71 | 24 - 97.96 |
| Mem Usage (MB-%)| 6.1 - 100.00 | 2 - 100.00 | 40.1 - 86.55 | 14 - 99.37 |
## [258. Add digits,][258]
This is one of those cases where you need to examine what the results are in order
to realize the solution. After running mucho input cases, notice that the result
is `num % 9` with special cases if `num == 0` and `num % 9 == 0`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 99.49 | 20 - 99.19 |
| Mem Usage (MB-%)| 5.7 - 39.50 | 2 - 100.00 | 39.9 - 84.01 | 14.2 - 36.59 |
## [263. Ugly Number.][263]
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 21 - 99.88 |
| Mem Usage (MB-%)| 5.4 - 97.68 | 2.1 - 61.11 | 2.1 - 6.30 | 14 - 12.36 |
## [268. Missing Number,][268]
Yet another variation of the xor trick. Since all numbers are in range `[0, n)` and
we know only one is missing, we can use another variable ranging from `[0, n)` and
xor it along with all the numbers in the numbers array.
Since numbers with the same value will be xored out, the missing number will be the
one left.
Runtime complexity is `O(N)`, space complexity is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 92.23 | 0 - 100.00 | 80 - 91.77 | 120 - 96.29 |
| Mem Usage (MB-%)| 6.5 - 75.34 | 2 - 97.30 | 41.3 - 34.19 | 15.3 - 82.34 |
## [278. First bad version.][278]
Binary search, again.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 96.69 | 24 - 93.25 |
| Mem Usage (MB-%)| 5.5 - 53.47 | 2 - 81.08 | 38.4 - 45.85 | 14.1 - 73.36 |
## [283. Move zeroes.][283]
In order to minimize the comparisons made, our `run` variable always begins scanning
from the position in which a non-zero value was last found (which is zero in the
beginning). This way we only see each element in the nums array twice (once for the
`on_zero` variable tracking zeroes and once for the `run` variable tracking non-zero
values).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 100.00 | 0 - 100.00 | 76 - 98.15 | 32 - 99.97 |
| Mem Usage (MB-%)| 7.6 - 89.22 | 2.1 - 97.87 | 40.3 - 51.81 | 15.2 - 68.67 |
## [292. Nim Game,][292]
After some experimenting, we can see that the losing starting position is a multiple
of `4`. This makes sense, if we are on a multiple of `4`, our opponent can always
make the total stones decrease by `4` when playing optimally. This will lead to a
final count of `4` with us having to play and not being able to win the game.
In all other positions we can do the same, taking the correct ammount initially in
order to leave our opponent to make their first move with a number of rocks that is
a multiple of `4`.
Space/Time complexity is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 89.79 | 20 - 98.68 |
| Mem Usage (MB-%)| 5.5 - 41.79 | 1.9 - 75.00 | 38.2 - 76.88 | 13.9 - 97.16 |
## [326. Powers of three.][326]
Similar to powers of two. Find `log3(n)` and then check how close `round(log3(n))` is to `log3(n)`. Powers of
three will have a value of `log3(n)` close to some integer `x`. I'm under the impression a better solution
exists but it is 4am and no.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 97.77 | 4 - 100.00 | 244 - 76.58 | 56 - 98.86 |
| Mem Usage (MB-%)| 6.5 - 13.41 | 2.6 - 5.00 | 49.4 - 43.63 | 14.4 - 13.70 |
## [338. Counting bits.][338]
Straight-forward dp solution. num of bits for `n` is equal to num of bits `n >> 1` plus one if the shifted
value was a `1`. We can use tabulation to hold the values and gradually fill it up. `O(N)` runtime.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 36 - 96.89 | 4 - 94.74 | 84 - 99.78 | 72 - 96.85 |
| Mem Usage (MB-%)| 10.4 - 86.65 | 2.4 - 100.00 | 44.5 - 77.83 | 20.9 - 40.74 |
## [342. Power of four.][342]
Exactly the same as Powers of three, only difference is that we use `log4(n)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 100.00 | 24 - 95.62 |
| Mem Usage (MB-%)| 5.9 - 12.62 | 2.7 - 5.56 | 40 - 36.68 | 14.3 - 6.03 |
## [344. Reverse String][344]
Typical swapping.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 44 - 96.89 | 16 - 94.12 | 100 - 96.48 | 184 - 96.43 |
| Mem Usage (MB-%)| 12.4 - 74.64 | 5.4 - 98.82 | 45.5 - 79.32 | 18.7 - 14.37 |
## [345. Reverse vowels of a string.][345]
Use `start` and `end` trackers and swap on encountering two vowels. Where mutating the
string is allowed (Rust, C), it's done. Overall `O(N)` runtime for all and `O(1)/O(N)`
for where we can mutate the string or not respectively.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 84 - 99.45 | 44 - 94.72 |
| Mem Usage (MB-%)| 6.5 - 100.00 | 2.5 - 90.91 | 43.2 - 99.59 | 15.1 - 71.47 |
## [347. Top k frequent elements.][347]
Counting + Building a k-element heap. Complexity of operation is `O(NlogK)` instead of
`O(NlogN)`. Python contains a built-in version of this `Counter.most_common` so Rust is
probably the best place to look for each of these steps layed out sequentially.
C: Not up for implementing my own binary heap (probably should and just keep it around).
Javascript: Priority Queue available seems somewhat broken to me? I can't seem to get it to
work for some reason.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 0 - 100.00 | | 88 - 98.51 |
| Mem Usage (MB-%)| | 2.4 - 65.63 | | 18.7 - 76.49 |
## [349. Intersection of two arrays.][349]
Build sets and get intersection (or write it for `C` and `JS` cases.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 98.18 | 32 - 99.40 |
| Mem Usage (MB-%)| 7.7 - 12.87 | 2.1 - 77.78 | 40.5 - 28.27 | 14.4 - 22.10 |
## [367. Valid perfect square.][367]
Use babylonian method to find a good approximation of the root and then check that
`approx * approx == num`. Using `root = pow(num, 0.5)` seemed like it was going against
the point.
Same notes as in problem 50 - Sqrt(x) apply.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 98.82 | 24 - 95.16 |
| Mem Usage (MB-%)| 5.4 - 64.94 | 1.9 - 81.82 | 38.8 - 12.48 | 14.4 - 6.53 |
## [374. Guess number higher or lower.][374]
Simple binary search for the number. `O(logn)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 88.26 | 20 - 98.89 |
| Mem Usage (MB-%)| 5.5 - 52.67 | 1.9 - 66.67 | 38.3 - 81.91 | 14.2 - 42.20 |
## [377. Combination sum IV.][377]
Tabulation solution looks magical, as always, so recursive solution in python file would probably
be a better place to start. Not sure about complexity, I'd need to think it through
longer (I should probably write a big list of DP problems I've solved.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.04 | 28 - 99.28 |
| Mem Usage (MB-%)| 5.8 - 64.71 | 2 - 100.00 | 40.7 - 21.67 | 14.1 - 95.74 |
## [383. Ransom Note][383]
Use a counter to check that the required ammount of characters are present.
In C, Rust and JavaScript, an array of size 26 is filled up with the characters in the
magazine and then the characters from the ransom note are subtracted. If at any point
the value inside the array is zero, it means we've run out of characters.
Runtime complexity is `O(N + M)` where `N` and `M` are the sizes of the strings given.
Space complexity is `O(1)` since we only have 26 characters.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 99.57 | 40 - 86.90 |
| Mem Usage (MB-%)| 6.4 - 96.59 | 2.1 - 100.00 | 41.2 - 91.14 | 14.1 - 99.69 |
## [384. Shuffle an Array.][384]
Uses Fisher-Yates algorithm (the Durstenfeld's version) in order to shuffle sequence. This
leads to `O(N)` execution.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 212 - 82.28 | 24 - 100.00 | 228 - 89.90 | 264 - 89.11 |
| Mem Usage (MB-%)| 36.9 - 29.54 | 5.4 - 60.00 | 51.5 - 96.63 | 19.4 - 71.11 |
## [387. Find Unique Character in String.][387]
Build a Counter (in Python we use library supplied, in others, just use an array with
26 slots to act as Counter) and then go through characters of `s` and find the first
for which the count is equal to `1`.
Runtime complexity is `O(N)` for all, space is `O(1)` I believe since we always have
a collection with 26 elements regardless of string length.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 99.00 | 0 - 100.00 | 96 - 95.44 | 88 - 90.37 |
| Mem Usage (MB-%)| 6.9 - 80.90 | 2 - 78.26 | 42.4 - 58.18 | 14.4 - 70.26 |
## [389. Find the difference.][389]
Xor trick, different problem statement.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 98.01 | 28 - 92.43 |
| Mem Usage (MB-%)| 5.7 - 96.92 | 2 - 100.00 | 39 - 96.03 | 14.1 - 94.66 |
## [392. Is subsequence.][392]
Continuously search in the string `t` for the characters of `s`. Use bounds to not search
from the beginning for every char of `s` (and result in `O(nm)` complexity). `O(n)` time
complexity and `O(1)` space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.25 | 24 - 95.69 |
| Mem Usage (MB-%)| 5.5 - 97.97 | 2 - 100.00 | 37.8 - 99.55 | 14.2 - 73.80 |
## [400. Nth digit.][400]
Pre-compute number of digits until `9, 99, 999, 9999, ..., 99999999` and store them
in an array. To find the k-th digit we then need to find where in that array `n` lies and
perform some simple arithmetic in order to find the number and digit that `n` denotes.
We grab the largest value in the array which is smaller than `n`. This sets our lower bound.
To find the number we just need to find out how many numbers are contained in `n - largest_value`
digits and what's left after that. That can be done with a pretty a simple divmod. All in all,
this is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.83 | 12 - 100.00 |
| Mem Usage (MB-%)| 5.7 - 50.00 | 2 - 60.00 | 37.9 - 100.00 | 14.2 -41.10 |
## [404. Sum of left leaves.][404]
Relatively straight-forward, recurse on tree and while branching on the children
check if the left child is a leaf. If so, add its value, if not, recurse on it.
On right, we immediately recurse.
All solutions have used recursion, translating into stack based solution is easy.
This visits each node once, so `O(N)` time and space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 99.26 | 28 - 89.99 |
| Mem Usage (MB-%)| 6.3 - 91.09 | 2.2 - 77.78 | 40.1 - 43.18 | 14.7 - 81.31 |
## [405. Convert a number to hexadecimal.][405]
Use a small table `[0, 1, 2, ..., 14, 15] => ['0', '1', '2', ..., 'e', 'f']` to translate between
the values from `[0-15]` to appropriate hex symbols. We manipulate the input number `n` in pairs of
`4`bits (`n & 15` and `n >> 4`) which we use to look up the value in the table.
To handle 2's complement (infinite 1's to the left), we simply bound the loop to generate a maximum of
`8` hex symbols.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 90.67 | 24 - 94.23 |
| Mem Usage (MB-%)| 5.4 - 85.25 | 2 - 100.00 | 38.4 - 90.67 | 14.3 - 11.14 |
## [409. Longest Palindrome.][409]
Build a counter out of the characters and then add all even counts and all odd minus one counts.
If we encounter an odd along the way, we add `1` (to be placed in the middle) to get longest
palindrome.
Runtime/memory complexity is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 95.79 | 28 - 89.22 |
| Mem Usage (MB-%)| 5.7 - 100.00 | 2 - 80.00 | 39.6 - 82.30 | 14.2 - 81.53 |
## [412. Fizz-buzz.][412]
Straight-forward translation of statement, only trick is not performing `i % 3` and
`i % 5` twice do see if a number is a multiple of both `5` and `3`. We can use the
result of `i / 3` and see if that is divided by `5`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 69.37 | 0 - 100.00 | 76 - 96.82 | 34 - 92.74 |
| Mem Usage (MB-%)| 7 - 86.04 | 2.6 - 95.24 | 41.4 - 5.36 | 15 - 73.21 |
## [429. N-ary Tree Level Order Traversal.][429]
Similar to problem [Binary Tree Level Order Traversal][102]. The only difference being
that we know iterate through all children and add them to the stack. (implicit or explicit one.)
Complexity is the same, `O(N)` runtime/memory.
Rust: not available.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | N/A | 92 - 94.98 | 44 - 95.28 |
| Mem Usage (MB-%)| - | N/A | 43.7 - 26.96 | 16 - 87.08 |
## [434. Number of segments in a string.][434]
Split string on whitespace and filter empty items on resulting iterable. C needs to
manually step through the string. `O(N)` time (and space).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 95.61 | 20 - 98.30 |
| Mem Usage (MB-%)| 5.4 - 96.43 | 2 - 100.00 | 38.1 - 94.74 | 14.1 - 65.77 |
## [448. Find all numbers disappeared in an array.][448]
Couldn't see (yet) how to do it with `O(1)` space. For now, I use `O(N)` space
in the form of a vector or set in order to find disappeared numbers.
*TODO: Revisit but not urgently.*
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 92 - 98.94 | 8 - 98.04 | 96 - 99.71 | 312 - 99.29 |
| Mem Usage (MB-%)| 18.4 - 18.62 | 2.6 - 43.14 | 47.4 - 47.02 | 24.7 - 17.49 |
## [451. Sort characters by frequency.][451]
Count the characters, sort them by count, build the string.
The `O(N)` traversal of the string dominates the `O(klogk)` of the sort
due to the second being applied to an `k` which has a maximum value
of `count(letters + digits)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 88 - 92.53 | 24 - 99.90 |
| Mem Usage (MB-%)| 6.5 - 78.57 | 2.2 - 93.75 | 41.3 - 88.19 | 15.6 - 39.61 |
## [460. Minimum Changes to make alternating binary string.][460]
Need to count changes both version of alternating strings (`'010101..', '101010..'`). Call
a counting function that finds the counts for each of these cases and return the min value. Since
we only iterate through `s`, runtime complexity is `O(N)`. Space complexity is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 92.66 | 40 - 92.84 |
| Mem Usage (MB-%)| 5.7 - 100.00 | 2 - 100.00 | 39.4 - 51.05 | 14.2 - 96.60 |
## [461. Hamming Distance][461]
Get the xor of the two numbers (where bits differ) and then
count set bits.
I'm under the impression I'm missing some other obvious
solution here.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 82.30 | 28 - 76.20 |
| Mem Usage (MB-%)| 5.5 - 67.52 | 2 - 100.00 | 38.5 - 71.22 | 14.1 - 61.18 |
## [476. Number Complement.][476]
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 85.35 | 24 - 94.21 |
| Mem Usage (MB-%)| 5.6 - 63.49 | 2 - 80.00 | 38.1 - 94.27 | 14 - 95.98 |
## [485. Max Consecutive Ones][485]
We have to scan the full array looking for the max consecutive number of ones. Runtime
complexity is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 36 - 95.44 | 0 - 100.00 | 80 - 92.82 | 332 - 96.94 |
| Mem Usage (MB-%)| 7.4 - 97.54 | 2 - 85.71 | 41,6 - 32.50 | 14.3 - 92.44 |
## [495. Teemo attacking.][495]
Just iterate through the timeseries and the seconds as we go. `O(N)` time complexity
and `O(1)` space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 28 - 100.00 | 4 - 100.00 | 72 - 100.00 | 240 - 93.39 |
| Mem Usage (MB-%)| 7.1 - 100.00 | 2.2 - 33.33 | 42.5 - 92.77 | 16.2 - 5.97 |
## [496. Next greater element I.][496]
Python has three solutions, brute force, with a bit of supporting structures and
with a lot. Last is used and in the others and described here.
We initialize the resulting array to all -1's and create a mapping from values to
their indices. We go through each element in `nums2` and for each:
1. If it is contained in our mapping, we add it to a set that keeps track of values
seen.
2. We iterate through the values of seen, if, for any value of seen, we find that
it is smaller than the element of `nums2` we're iterating through, we add it
to the result array (where we add it is found by using the mapping).
For any values not in `nums1` and for any values in `nums1` for which no larger
element exists on the right, the initial value of `-1` stays. The trick here is
that for each element of `nums2` we only iterate over a very small subset of
`nums1` (the ones in `seen`). This way we can reduce the overall iterations
performed.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 97.12 | 36 - 98.72 |
| Mem Usage (MB-%)| 10.4 - 6.67 | 2 - 100.00 | 40.5 - 42.45 | 14.6 - 18.78 |
## [500. Keyboard Row][500]
Make sets out of rows on keyboard and check if input strings are contained
in the sets.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 97.03 | 24 - 92.53 |
| Mem Usage (MB-%)| 5.8 - 20.69 | 2 - 100.00 | 38.6 - 33.83 | 14.1 - 44.11 |
## [504. Base 7.][504]
Find the digits by continuously taking `num mod 7` and reducing `num` by `/ 7`. This is done on
the absolute value of the number, the sign is removed and its presence is tracked in a boolean flag.
After that, we only need to build the resulting string.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 99.32 | 24 - 95.12 |
| Mem Usage (MB-%)| 5.6 - 75.00 | 2 - 85.71 | 39.3 - 6.12 | 14.3 - 43.06 |
## [509. Fibonacci Number][509]
Calculate it iteratively.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 97.37 | 20 - 98.15 |
| Mem Usage (MB-%)| 5.4 - 86.96 | 2 - 48.78 | 38.3 - 56.54 | 14.1 - 34.89 |
## [515. Find largest value in each tree row.][515]
Straight-forward. Keep an array of maxes and recurse/iterate while holding on to
a variable denoting the level. Update the maxes array using that level. Uses DFS and
visits all nodes. `O(N)` space and time complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 80 - 98.94 | 40 - 93.53 |
| Mem Usage (MB-%)| 14.7 - 8.33 | 3 - 100.00 | 44.2 - 8.80 | 16.5 - 57.81 |
## [520. Detect Capital.][520]
We can just check the second character and disambiguate which case we're examining.
Then we just make sure all the following characters match the case.
`O(N)` runtime complexity with `O(1)` space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 100.00 | 20 - 98.83 |
| Mem Usage (MB-%)| 5.5 - 98.33 | 2 - 60.00 | 39.8 - 81.21 | 14.2 - 48.87 |
## [521. Longest Uncommon Subsequence.][521]
If the strings are equal, all subsequences are the same, return -1.
Else, return the largest among the strings.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 95.35 | 24 - 93.63 |
| Mem Usage (MB-%)| 5.4 - 100.00 | 2 - 100.00 | 38.4 - 61.63 | 14 - 98.24 |
## [530. Minimum absolute difference in bst.][530]
C and Rust both traverse the bst and build a sorted vector. Python and Javascript
use a generator to yield back values. Whatever the case, we then traverse through
the value and find the minimum.
Note: Javascript would probably be faster if I didn't use generators. Problem is,
I really like generators so I don't care.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 97.44 | 0 - 100.00 | 128 - 14.84 | 48 - 96.33 |
| Mem Usage (MB-%)| 10.8 - 20.51 | 2.9 - 25.00 | 46.2 - 9.03 | 16.3 - 32.45 |
## [535. Encode and Decode TinyURL.][535]
As basic as it gets implementation. (Note: just returning the string as-is actually
works, they don't test to guard against this).
There's an issue with Rust in that the test code declares the coder as immutable
thereby not allowing us to pass it mutably by reference.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | N/A | 84 - 93.40 | 24 - 97.30 |
| Mem Usage (MB-%)| | N/A | 40 - 83.96 | 14.2 - 58.76 |
## [538. Convert BST to Greater Tree][538]
Code used is exactly the same for [Binary Search Tree to Greater Sum Tree](1038).
As such, see `1038.*` files for the code.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 92.86 | 0 - 100.00 | 100 - 100.00 | 72 - 98.64 |
| Mem Usage (MB-%)| 13.7 - 100.00 | 2.9 - 83.33 | 47.7 - 28.72 | 16.7 - 80.07 |
## [539. Minimum Time Difference.][539]
Idea is transform times into a continuous range of ints (i.e 0 - 1440) by multiplying hours
by `60`. After doing that we can go through transformed values and find minimum. Since time wraps
around we need to check if the maximum value and the minimum value are closer than the minimum found
after traversing array.
Using a set to check for duplicates (and return early if one is found) appears to be a good optimization.
Overall, this uses `O(N)` additional space and runs in `O(NlogN)` due to the sorting.
Note for C: This can easily be translated there. Dealing with sets is a pain though and I'm not into it
anymore.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 0 - 100.00 | 84 - 99.11 | 52 - 99.37 |
| Mem Usage (MB-%)| | 3.1 - 100.00 | 41.6 - 62.83 | 16.9 - 92.87 |
## [547. Number of provinces.][547]
Dfs on nodes. Adjacency matrix representation doesn't help at all with this but meh. We still traverse
O(V**2) times.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 24 - 80.00 | 0 - 100.00 | 72 - 97.31 | 180 - 91.43 |
| Mem Usage (MB-%)| 7 - 50.00 | 2.1 - 100.00 | 39.9 - 98.59 | 14.7 - 72.30 |
## [551. Student Attendance Record I.][551]
Iterate through characters with two counters. One counts occurences of `A`s and returns
if the count reaches `2`. The second counts consecutive occurences of `L`s and returns if
it equals `3`. Overall, `O(N)` runtime and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 93.62 | 20 - 99.19 |
| Mem Usage (MB-%)| 5.5 - 73.81 | 2 - 77.78 | 38.8 - 41.49 | 14 - 90.13 |
## [557. Reverse Words in a String III][561]
C mutates the string in-place because it's allowable there.
The rest split on `' '`, iterate through the chunks and
reverse them.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 93.81 | 0 - 100.00 | 88 - 74.93 | 24 - 97.41 |
| Mem Usage (MB-%)| 6.9 - 72.57 | 2.3 - 72.73 | 44.6 - 85.18 | 14.6 - 84.18 |
## [559. Maximum depth of N-Ary Tree][559]
Common recursive approach (call depth on all children.)
Note: For some reason, `maxDepth` in `C` is defined as returning `int *`.
Note: Rust case not applicable, they don't allow the option yet.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | N/A | 84 - 95.55 | 40 - 91.04 |
| Mem Usage (MB-%)| 7.3 - 8.70 | N/A | 42.1 - 33.27 | 16.2 - 7.06 |
## [561. Array Partition I][561]
Sort the array in reverse order and sum every second
element.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 52 - 80.25 | 8 - 96.15 | 120 - 93.30 | 244 - 97.34 |
| Mem Usage (MB-%)| 8.5 - 27.16 | 2.2 - 100.00 | 44 - 79.12 | 16.5 - 81.43 |
## [566. Reshape the matrix.][566]
Check that new dimensions can hold elements and then go through each
element in `nums` and place it in new position based on `r` and `c`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 36 - 100.00 | 0 - 100.00 | 96 - 96.50 | 84 - 99.31 |
| Mem Usage (MB-%)| 11.4 - 100.00 | 2.3 - 100.00 | 44.6 - 51.05 | 15.2 - 59.61 |
## [572. Subtree of another tree.][572]
Traverse tree until a node with value equal to `subRoot.val` is encountered. Then check if
subtrees match using recursion. This could probably be improved by building a list of the
`subRoot` values and checking the nodes against that.
Rust: Need to come back to this, refcells with references are still difficult for me.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 99.17 | - | 92 - 98.19 | 132 - 81.28 |
| Mem Usage (MB-%)| 10.9 - 98.33 | - | 44.9 - 90.86 | 15 - 88.52 |
## [575. Distribute Candies][575]
Use a set and compare its size with the ammount prescribed (Goes south for `C`
though, which might mean there's a different better way at it.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 224 - 21.43 | 24 - 100.00 | 128 - 90.75 | 764 - 92.56 |
| Mem Usage (MB-%)| 68.3 - 7.14 | 2.2 - 86.67 | 52.7 - 35.24 | 16.5 - 8.60 |
## [589. N-Ary tree pre-order traversal][589]
Option for Rust isn't available yet unfortunately.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 82.00 | N/A | 92 - 89.63 | 48 - 81.31 |
| Mem Usage (MB-%)| 19.9 - 24.00 | N/A | 43.1 - 13.39 | 15.6 - 86.89 |
## [590. N-Ary tree post order traversal][590]
All using iterative versions (probably partly explains why
timings show sluggishness).
Option for Rust isn't available yet unfortunately.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 14.63 | N/A | 100 - 55.56 | 32 - 100.00 |
| Mem Usage (MB-%)| 24.4 - 9.76 | N/A | 43.7 - 9.69 | 15.8 - 57.01 |
## [605. Can place flowers.][605]
In short, we need to go through the array in three pairs `i-1, i, i+1` and check
if we can place a flower there. Main trick we can utilize is to increment `i` by
more than one in specific cases.
See Python file for step by step comments on what is done. Overall, `O(N)` (eh, `N/2`)
in the worse case.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 98.04 | 0 - 100.00 | 80 - 90.99 | 148 - 99.49 |
| Mem Usage (MB-%)| 7 - 100.00 | 2.1 - 100.00 | 40.5 - 79.19 | 14.6 - 70.07 |
## [606. Construct string from binary tree.][606]
Preorder traversal through the tree and build the string. If we don't have a left
subchild but do have a right one, we need to add the extra `'()'` as stated in
the problem statement.
For C, couldn't be buggered, really. Need to recurse through the tree with a
pointer to `char *` array and add things to it. Then, join them all together
at the end. Might do some point in the future, for now, nope.
Time complexity is `O(N)`, space complexity is probably the same.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 84 - 97.62 | 40 - 98.24 |
| Mem Usage (MB-%)| - | 3.1 - 100.00 | 48.6 - 5.56 | 16.5 - 10.29 |
## [617. Merge Two Binary Trees][617]
Don't recurse. Breadth first traversal using a queue.
Add branches immediately when we can.
Pending:
- `C` because I need to make a small queue first.
- `Rust`: Got to sit down and learn `Rc`s and `RefCell`s better. [OK, more familiar now].
- `Javascript`: File added, timings need to improve.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 4 - 100.00 | | 80 - 96.68 |
| Mem Usage (MB-%)| | 2.2 - 100.00 | | 14.9 - 99.93 |
## [637. Average of Levels in Binary Tree][637]
Python, Rust use iterative approach: we use a deque that holds a special
marker (i.e `None`) which we use to track on which level we are on. Until we
reach that marker, we add the values and count the nodes. When we reach the
marker, we take the average and add it to the result array.
The iteration stops when for a given level, the number of Nodes we've counted
is zero.
`C` and `Javascript` use a recursive approach whereby each level of the tree
is represented by a `level` parameter of the recursive function.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 84 - 97.80 | 36 - 99.85 |
| Mem Usage (MB-%)| 20.7 - 5.36 | 3 - 50.00 | 45 - 18.13 | 16.5 - 60.71 |
## [641. Design Circular Deque.][641]
Two common approaches for this: a doubly linked list (see Python) or a circular
buffer/list (see C/Rust/Javascript). The main difference is in memory requirements since
a doubly linked list needs to store references to front/back for all nodes.
All in all, memory footprint for both is `O(N)` and both display `O(1)` complexity for
all operations.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 100.00 | 4 - 100.00 | 104 - 100.00 | 64 - 91.50 |
| Mem Usage (MB-%)| 12.5 - 80.00 | 2.4 - 66.67 | 45.9 - 77.14 | 15.1 - 43.28 |
## [643. Maximum average subarray I.][643]
`O(N)`, the main thing to notice (I believe) is the fact that you do not need to calculate the
full summation for the average each time (k terms added and divided by k). Instead, you can build
the summation until `k` and then move along the array looking for the max. At the end of each loop
we remove the first term of the summation and add the term we've currently looked at.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 100 - 100.00 | 12 - 100.00 | 92 - 97.89 | 780 - 96.34 |
| Mem Usage (MB-%)| 10.5 - 50.75 | 2.4 - 33.33 | 47.9 - 37.19 | 18.1 - 10.88 |
## [645. Set Mismatch.][645]
My solution is in Python where a Set and the sum until `n` is used in order to find these.
Did peek at solution that plays around with the indices and sets them to negatives. This is a solid
trick to remember whenever dealing with arrays holding sequences that can be used to index it (or
holding sequences of values that can be transformed into the range `0..array.len()`).
First approach with Sets uses `O(n)` runtime complexity and `O(n)` space. Second approach uses two
iterations but `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 97.47 | 0 - 100.00 | 84 - 94.77 | 192 - 71.13 |
| Mem Usage (MB-%)| 7.1 - 79.75| 2.1 - 100.00 | 42.2 - 68.11 | 15.2 - 97.92 |
## [653. Two Sum IV - Input is a BST.][653]
Use a set to hold the values and traverse tree. For each node see if `k - node.value`
is present inside the set, if so, done.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 98.11 | 0 - 100.00 | 104 - 97.44 | 68 - 97.46 |
| Mem Usage (MB-%)| 25.6 - 18.87 | 3.3 - 80.00 | 48.7 - 22.16 | 16.3 - 94.07 |
## [654. Maximum Binary Tree.][654]
Work with ranges. Start by calling recursive function in full range. Find max and its
index in the nums array and then recurse on `[start, id_max]` for the left subtree
and `[id_max + 1, end]` for the right subtree.
Create node and return.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 36 - 98.11 | 8 - 100.00 | 108 - 96.08 | 192 - 88.87 |
| Mem Usage (MB-%)| 26.9 - 90.57 | 2.1 - 100.00 | 45.1 - 77.94 | 15.1 - 8.05 |
## [657. Robot Return to Origin][657]
Count occurences and compare. If number of 'D's matches number
of 'U's and number of 'L's matches number of 'R's, we're good.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 95.76 | 0 - 100.00 | 80 - 90.34 | 32 - 94.10 |
| Mem Usage (MB-%)| 6 - 57.63 | 1.9 - 100.00 | 39.7 - 61.54 | 14.2 - 67.31 |
## [669. Trim a binary Search Tree.][669]
Initially we need to see where our root value lies. If it isn't inside the
acceptable range, we need to adjust it until it is. This is done by setting
the root to right or left depending if the value of root is smaller than
`low` or larger than `high` respectively.
After we have found a new root, we continuously recurse on children and, by
keeping a reference to their parent, trim any node not inside the range by
re-assigning pointers/attributes. Recursion is terminated when we reach
all leaves.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 76.47 | 0 - 100.00 | 88 - 86.70 | 40 - 97.47 |
| Mem Usage (MB-%)| 10.6 - 64.71 | 2.9 - 87.50 | 44.2 - 83.74 | 14.1 - 46.08 |
## [674. Longest continuous increasing subsequence.][674]
Iterate through array and keep track of longest sequence encountered. `O(N)` runtime
and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 87.79 | 68 - 93.20 |
| Mem Usage (MB-%)| 6.4 - 65.67 | 2.3 - 80.00 | 39 - 95.35 | 15.4 - 57.78 |
## [680. Valid Palindrome II.][680]
Start typically by comparing characters via two counters, start and end. When a different character
is encountered, compare the substrings `s[start+1:end]` and its reverse, if that isn't
equal compare `s[start:end-1]` with its reverse. Essentially, skip the character for the two cases
that appear (either from start, or from end). Overall, `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 91.53 | 0 - 100.00 | 13 - 96.03 | 94 - 97.78 |
| Mem Usage (MB-%)| 9.3 - 33.90 | 2.4 - 34.48 | 10 - 5.05 | 14.5 - 49.54 |
## [682. Baseball Game][682]
While loop your way through the operations, make sure you skip performing an
operation if it is followed by a `"C"`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 92.51 | 32 - 96.73|
| Mem Usage (MB-%)| 6.2 - 55.17 | 2.1 - 100.00 | 38.9 - 90.64 | 14.4 - 53.20 |
## [684. Reduntant Connection.][684]
Use a disjoint set union structure and track cycles (where `find(a) == find(b)`). Iterate
through all edges and hold on to the last one that would create a cycle. Overall, this
runs in `O(E)` and has memory requirements of the same order.
Alternatively, we could transform into adjacency list representation, run dfs and
check for back-edges. Since we get and edge list, the union-find solution is
probably better.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 93.97 | 52 - 88.08 |
| Mem Usage (MB-%)| 6.2 - 61.06 | 2.1 - 89.47 | 40.8 - 80.98 | 14.7 - 95.83 |
## [690. Employee Importance][690]
Transform employees into a dictionary keyed by the id. Then we can build a list of
all subordinates by continuously keying the dictionary and build up the overall
importance.
*NOTE: Rust and C versions not available.*
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| N/A | N/A | 80 - 97.80 | 144 - 98.93 |
| Mem Usage (MB-%)| N/A | N/A | 44.2 - 17.56 | 15.5 - 53.57 |
## [693. Binary Number with alternating bits.][693]
Check adjacent bits until we reach end. If at any point a pair of adjacent bits is the
same, return False.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 95.92 | 20 - 99.27 |
| Mem Usage (MB-%)| 5.5 - 55.67 | 1.9 - 66.67 | 38.6 - 53.06 | 14.4 - 5.57 |
## [696. Count binary substrings.][696]
Go through each character and build count of adjacent groups by using a couple
of flags. Count of adjacent groups can then be found easily.
Note: Rust :(
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 93.75 | 4 - 60.00 | 84 - 93.42 | 112 - 96.53 |
| Mem Usage (MB-%)| 6.7 - 100.00 | 2 - 100.00 | 41.8 - 92.11 | 14.5 - 88.91 |
## [697. Degree of an Array.][697]
Build a counter of the numbers in the array in order to find the maximum degree
along with which numbers have that maximum degree.
After that we can iterate through the numbers array from the left (start) and
right (end) and look for the leftmost and rightmost occurences for each of the
numbers that has that maximum degree.
The minimum difference `end - start` for each of the numbers with max degree is
the result.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 100.00 | 4 - 100.00 | 84 - 98.22 | 196 - 99.93 |
| Mem Usage (MB-%)| 9.8 - 22.22 | 2.2 - 100.00 | 44.5 - 36.89 | 15.2 - 98.41 |
## [700. Search in a BST][700]
Straight-forward. Perform binary search. `C` code also contains
a tail-recursive version, didn't seem to make much difference
in runtime/memory timings.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 89.04 | 4 - 100.00 | 44 - 98.93 | 64 - 97.56|
| Mem Usage (MB-%)| 15.1 - 73.25 | 2.6 - 57.14 | 44.4 - 76.63 | 15.8 - 94.95|
## [704. Binary Search.][704]
Well, binary search. `O(logn)` complexity, `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 28 - 98.79 | 0 - 100.00 | 76 - 92.07 | 216 - 99.84 |
| Mem Usage (MB-%)| 6.9 - 99.19 | 2.1 - 100.00 | 42.4 - 52.53 | 15.5 - 91.95 |
## [705. Design HashSet][705]
Python contains implementations for both separate chaining and open addressing, so
does `Rust` (well, an attempt at separate chaining, at least).
Have some odd bug in Javascript which I'll give up on trying to track down for now.
Adding code so far.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 76 - 100.00 | 16 - 100.00 | N/A | 196 - 57.80 |
| Mem Usage (MB-%)| 26.1 - 96.83 | 5.6 - 66 | N/A | 18.9 - 53.31 |
## [709. To lower case][709]
The `C` version is probably the most interesting. The rest
just use the language provided conversion methods.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 53.05 | 20 - 97.94|
| Mem Usage (MB-%)| 5.7 - 20.65 | 2 - 100.00 | 38.4 - 8.75 |14.1 - 99.88|
## [728. Self Dividing Numbers][728]
Go through digits of number and check for divisibility. Basically
brute-forcing our way to an answer. Not sure if anything better
than brute-force is possible.
Sidenote: Transforming to strings could be a better approach,
not sure.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 84 - 65.67 | 44 - 81.10 |
| Mem Usage (MB-%)| 6.2 - 14.75 | 2.2 - 100.00 | 38.8 - 11.69 | 14.1 - 100.00 |
## [744. Find smallest letter greater than target.][744]
We can immediately handle the wrap-around case by checking if `target` is `>=` than
the last element in the sorted vector. If that is the case, we immediately return the
first element.
If not, just traverse the array and find the character.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 96.63 | 100 - 94.06 |
| Mem Usage (MB-%)| 6 - 64.29 | 2.7 - 40.00 | 41.2 - 12.36 | 14.3 - 84.93 |
## [746. Min cost climbing stairs.][746]
Dynamic Programming with tabulation. Solutions for step `n` is the minimum between the solutions at
one step `n-1` and the two steps `n-2` back, plus the cost of the step at `n`.
`O(N)` runtime and space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 97.35 | 48 - 97.19 |
| Mem Usage (MB-%)| 5.9 - 60.49 | 0 - 100.00 | 39.8 - 56.16 | 14.1 - 99.68 |
## [748. Shortest completing word][748]
To be frank, don't think things are good here. Might need to place this problem in
the revisited section. I won't much bother with `C` with the current solution that
uses a host of maps/sets.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 4 - 66.67 | 96 - 75.76 | 60 - 97.43 |
| Mem Usage (MB-%)| | 2.1 - 66.67 | 43.7 - 51.51 | 14.6 - 24.70 |
## [766. Toeplitz Matrix][766]
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 98.44 | 0 - 100.00 | 84 - 91.64 | 80 - 86.38 |
| Mem Usage (MB-%)| 6 - 98.44 | 2 - 75.00 | 40.5 - 40.68 | 14.1 - 90.26 |
## [771. Jewels and Stones][771]
Add the values of J to a set and count number of jewels by
using the set membership test to quickly establish if a
character is a jewel.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 73.22 | 24 - 93.75 |
| Mem Usage (MB-%)| 5.8 - 25.51 | 2.1 - 100.00 | 40.4 - 11.66 |14.2 - 99.99 |
## [783. Minimum Distance between bst nodes.][783]
Same as problem [530. Minimum absolute difference in bst][530], see there.
## [788. Rotated Digits][788]
I'll probably need to write this up in a separate `.md` file. Until then,
see the source code for the Python case where I prototyped and added
most comments.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 94.16 | 20 - 99.86 |
| Mem Usage (MB-%)| 5.7 - 45.71 | 2 - 66.67 | 38 - 100.00 | 14.1 - 90.83 |
## [797. All paths from source to target.][797]
Perform a dfs without tracking visited nodes. Since we have a DAG, we needn't worry about
recursing infinitely. We just need to incrementally build a path indicating our current
visit from the source and keep track of it when we encounter our destination.
Python contains both recursive and iterative, Rust and Javascript contain iterative. C is
similar but, once again, manual memory allocation.
This follows every single edge so it should be `O(E)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 8 - 100.00 | 100 - 97.30 | 92 - 96.32 |
| Mem Usage (MB-%)| - | 2.7 - 100.00 | 46.6 - 52.13 | 15.5 - 91.83 |
## [804. Unique Morse Code Words.][804]
Build the morse code words and add them to a set. After adding
all words just return the length of the, now, unique set of
morse code words.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 86.43 | 24 - 99.14|
| Mem usage (mb-%)| 6.1 - 13.33 | 2 - 100.00 | 39.4 - 6.79 | 14.1 - 100.00 |
## [806. Number of lines to write string.][806]
Build a mapping from lowercase english characters to their width in the `widths`
array. Then, gradually find the lines needed.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 90.28 | 20 - 99.40 |
| Mem usage (mb-%)| 5.8 - 44.44 | 2 - 66.67 | 39.4 - 12.50 | 14.2 - 48.64 |
## [807. Max increase to keep city skyline.][807]
Max ammount that we can grow is `min(max_of_row, max_of_col)`. We can just find
these values and compute the maximum increase.
Pretty sure complexity is around `O(N*M)`, inner loop computing maximum for column only
runs once per column, this means that we have a significant lower term in our
complexity equation.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 93.02 | 0 - 100.00 | 80 - 91.76 | 64 - 96.36 |
| Mem usage (mb-%)| 6 - 97.67 | 2.1 - 86.67 | 39.1 - 92.86 | 14.3 - 61.22 |
## [811. Subdomain visit count.][811]
Hold domains/subdomains in a Counter and count values after
splitting domain up.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 71.43 | 4 - 100.00 | 92 - 93.28 | 44 - 95.94|
| Mem usage (mb-%)| 9.6 - 28.57 | 2.2 - 25.00 | 43.4 - 70.71 | 14.4 - 26.05 |
## [812. Largest Triangle Area.][812]
Initially I tried to see if we can skip taking all combinations by trying to find
two points that are maximally appart and using those as the two initial points.
Unfortunately, this isn't correct.
After this I tried using Heron's formula to find the area based on the length of the
sides of the trianges (found using Euclidean distance). While working, this wasn't
very efficient.
Thankfully, after some digging around was able to find the Shoelace formula which made
my life easier because I only uses the three points of the triangle to calculate the
area.
Still, algorithm is dominated by the fact that we have to go through all combinations
(and I can't think of a way to reduce the iteration space).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 87.50 | 0 - 100.00 | 80 - 93.33 | 96 - 94.12 |
| Mem usage (mb-%)| 5.8 - 100.00 | 2 - 100.00 | 39 - 60.00 | 14.3 - 48.90 |
## [821. Shortest Distance to a Character.][821]
By keeping around the positions the character is found and the previous position
found, we can easily build the ranges after taking care of some special
cases.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 99.41 | 28 - 98.30 |
| Mem usage (mb-%)| 6 - 100.00 | 2 - 100.00 | 41 - 28.99 | 14.1 - 97.67 |
## [824. Goat Latin.][824]
Not much to break down here. Just do what the problem states.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 94.85 | 24 - 93.52 |
| Mem usage (mb-%)| 5.9 - 84.21 | 2 - 87.50 | 38.8 - 59.56 | 13.9 - 97.34 |
## [832. Flipping an Image][832]
Go through matrix and reverse/invert. Python/Rust/JS use a functional
approach. C iterates through elements.
**TODO: Write version for C where pointer arithmetic is used.**
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 67.03 | 0 - 100.00 | 80 - 86.11 | 40 - 99.08|
| Mem usage (mb-%)| 7.2 - 72.54 | 2.1 - 100.00 | 40.2 - 18.30 | 13.9 - 100.00 |
## [841. Keys and rooms.][841]
DFS from room zero and check if we can reach all rooms (# of visited == # of rooms). `O(N)` space and runtime
where `N` is total number of rooms.
Note: C is the same, manual allocation makes me too bored to implement now.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 76 - 91.87 | 60 - 92.21 |
| Mem usage (mb-%)| - | 2.1 - 100.00 | 40.2 - 79.95 | 14.8 - 81.62 |
## [845. Longest Mountain in Array.][845]
Step through array and attempt to perform an up-down motion through the values. We need
at least one up and one down step in order to register a mountain, in other situations we
don't have one.
When we find a mountain we advance the counter by the number of `up + down - 1` steps performed. This way
we can check if the end of the `down` part of a mountain starts the `up` part of another mountain.
This uses no extra storage and performs `O(N)` loops.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 100.00 | 0 - 100.00 | 64 - 100.00 | 156 - 94.06 |
| Mem usage (mb-%)| 6.6 - 90.91 | 2.1 - 100.00 | 41.1 - 40.82 | 15.1 - 99.73 |
## [852. Peak index in a Mountain Array.][852]
A binary search throug the semi-sorted array.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 89.84 | 0 - 100.00 | 72 - 91.56 | 64 - 97.11 |
| Mem usage (mb-%)| 6.6 - 30.47 | 2.1 - 75.00 | 39.1 - 77.91 | 15 - 88.20 |
## [859. Buddy strings.][859]
See comments in python file (`859.py`) for a better description. `O(N)` execution.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 100.00 | 24 - 98.49 |
| Mem usage (mb-%)| 5.7 - 56.92 | 2.2 - 81.82 | 41.2 - 35.74 | 14.5 - 71.61 |
## [867. Transpose Matrix.][867]
Straight-forward transpose.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 24 - 97.44 | 0 - 100.00 | 80 - 97.45 | 68 - 96.25 |
| Mem usage (mb-%)| 10.1 - 20.51 | 2.2 - 50.00 | 40.7 - 79.27 | 14.9 - 44.26 |
## [868. Binary Gap.][868]
Go through each bit of `n` (via `n >>= 1`) and count the distance between
two ones.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 97.12 | 20 - 98.78 |
| Mem usage (mb-%)| 5.6 - 46.88 | 2 - 100.00 | 38.8 - 50.00 | 14.2 - 57.69 |
## [872. Leaf similar trees.][872]
For Python: recursion delegating to sub-generators. Helpful.
For the rest: function that receives stack and returns leaves as we encounter
them.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 97.33 | 20 - 99.00 |
| Mem usage (mb-%)| 6.3 - 90.20 | 2 - 66.67 | 40.3 - 73.66 | 14.3 - 17.11 |
## [876. Middle of the Linked List][876]
Traverse the list while keeping a reference to the middle element.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 89.27 | 20 - 98.87 |
| Mem usage (mb-%)| 5.8 - 68.20 | 2 - 80.00 | 38.2 - 89.15 | 14 - 73.76 |
## [883. Projection Area of 3d Shapes][883]
Length of rows is the same.
View from Z axis is simply a count of all non-zero values.
View from X axis is the max of each row/cell.
View from Y axis is the max of each column.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 100.00 | 0 - 100.00 | 68 - 100.00 | 60 - 97.64 |
| Mem usage (mb-%)| 6 - 100.00 | 2 - 50.00 | 39 - 58.82 | 14.1 - 77.88 |
## [884. Uncommon words from two sentences.][884]
Create two sets for each String, a set of seen words and a set of words
unique in each String.
Then we take the difference of the unique words in string `A` and the
seen words in `B` (to only keep unique words not in the other string). We
do the same thing for the unique words of `B`.
Then we can take their union.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.65 | 24 - 94.61|
| Mem usage (mb-%)| 7.3 - 16.67 | 2.1 - 66.67 | 39.1 - 33.97 | 14.1 - 88.96 |
## [888. Fair Candy Swap.][888]
Brute force approach is going through each pair and checking.
A sligtly better option is to filter values based on if they bring us in the right
range (which can be found by adding the sums and dividing by two).
The best approach in the end is to use a set for one of the arrays. Then, knowing what
the difference between the two sums is in the beginning, we can go through the elements
of one array and find what the value should be in order to fill the difference between
the initial sums. If the value found is inside the set created for the other array, we've
found our element (which is guaranteed to exist).
**NOTE: C timings indicate either that uthash is slow or that a better option exists
that I haven't seen yet.**
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 88 - 48.16 | 8 - 100.00 | 100 - 97.93 | 332 - 99.56 |
| Mem usage (mb-%)| 26.6 - 11.11 | 2.5 - 100.00 | 48.1 - 28.28 | 16.6 - 41.79 |
## [896. Monotonic Array.][896]
Note that equal elements can't be ignored.
Find first non-equal pair in array, this initializes a flag that indicates the
direction the rest of the elements should be in in order for the array to be
monotonic. As soon as a pair isn't in the right direction, return `false`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 56 - 94.29 | 4 - 100.00 | 88 - 93.66 | 448 - 91.01 |
| Mem usage (mb-%)| 9.4 - 100.00 | 2.4 - 92.31 | 45 - 61.09 | 20.5 - 53.08 |
## [897. Increasing Order Search Tree][897]
Inorder traversal to gather nodes and then re-attaching
accordingly.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 98.18 | 24 - 94.91|
| Mem usage (mb-%)| 6.2 - 59.26 | 2.2 - 100.00 | 38.9 - 61.82 | 14 - 77.72 |
## [905. Sort Array By Parity.][905]
The two ways I've done this is either by simply sorting with the
appropriate key function or by using a deque.
Sorting takes `O(nlogn)` and `O(1)` space while using a deque
results in `O(n)` runtime complexity and `O(n)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 28 - 77.42 | 0 - 100.00 | 96 - 72.31 | 60 - 99.98|
| Mem Usage (MB-%)| 9.3 - 97.35 | 2.1 - 100.00 | 40.6 - 10.20 | 14.6 - 36.39|
## [908. Smallest Range I.][908]
Find min and max of the array. IF `max - K - min` falls in the range `[-K, K]` we
can use any value of that range to reduce the difference to `0`. If it doesn't,
then, if `max - K - min` is positive, subtract `K` from it (to minimize the
difference) or, if negative, add `K` to it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 88.89 | 0 - 100.00 | 72 - 96.74 | 96 - 100.00 |
| Mem Usage (MB-%)| 6.5 - 100.00 | 2 - 100.00 | 39.8 - 85.87 | 15.2 - 97.14 |
## [917. Reverse only letters.][917]
Code was written very quickly. Uses a set of acceptable characters and only switches when
we are on a valid pair of characters (in ascii letter range).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 94.95 | 24 - 95.12 |
| Mem Usage (MB-%)| 5.9 - 26.32 | 2 - 50.00 | 38.9 - 51.51 | 14.4 - 12.88 |
## [922. Sort Array By Parity II][922]
Preallocate resulting array. Then go through input array and using two
variables to denote the positions of the odds and evens, fill the
array.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 64 - 93.81 | 8 - 100.00 | 88 - 100.00 | 196 - 92.39 |
| Mem Usage (MB-%)| 13.5 - 51.55 | 2.1 - 77.78 | 44.4 - 74.18 | 16.8 - 15.40 |
## [929. Unique Email Address.][929]
Either step through the characters directly (see `C` and `Python` and `Rust` files)
or use string facilities that do that for you (see `Python` and `Javascript`).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 98.31 | 0 - 100.00 | 88 - 91.97 | 44 - 91.99 |
| Mem Usage (MB-%)| 8.1 - 10.17 | 2.1 - 93.33 | 44 - 41.89 | 14.4 - 7.90 |
## [933. Number of Recent Calls][933]
Keep t's in a Queue/Deque and using that find the correct
amount of pings. Keeping the last ping around in a
separate variable allows us to quickly answer with 1 and
clear the Queue/Deque.
**TODO: Need to implement a Queue for C for it to run. Code
is currently written as if using a Queue.**
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 32 - 94.22 | 228 - 72.72 | 260 - 99.58 |
| Mem Usage (MB-%)| | 5.6 - 84.39 | 50.2 - 29.84 | 18.7 - 91.13|
## [937. Reorder Data in log files.][937]
Separate into two arrays, one containing the logs that only have digits after the
identifier and one with the rest (containing letters).
The digits array doesn't need any processing.
The letters array is sorted based on the string after the identifier and, if they
compare equal, the identifier is used as a tie-braker (see C version for clear
illustration of this).
Splitting the arrays is `O(N)` while sorting the letters is `O(logM)` where `M`
is the total number of rows in the logs array that are letter based. Auxiliary
memory is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 90.00 | 0 - 100.00 | 80 - 95.48 | 28 - 97.50 |
| Mem Usage (MB-%)| 7.6 - 36.00 | 2.1 - 44.44 | 41.2 - 83.15 | 14.3 - 87.55 |
## [938. Range Sum of BST][938]
**TODO: Improve Rust, don't know Ref/RefCell good enough
to readjust root correctly. I've added a minimal
Tree impl to work with in playground.**
Do an inorder traversal and sum the items. We can
cut execution time considerably by adjusting the root
before we begin the traversal.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 72 - 99.00 | 12 - 100.00 | 204 - 98.44 |184 - 99.93|
| Mem Usage (MB-%)| 42.8 - 20.00 | 4.3 - 100.00 | 67.7 - 11.10 |22.3 - 99.95|
## [942. DI String Match][942]
Initialize counters, i to zero and j to the length of the string.
Iterate through characters of string and add i if character is
'I' and j if char is 'D'. Increase and decrease the counters
respectively afterwards.
Append either i or j at the end (they should be equal) to
complete the sequence.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 32 - 98.51 | 4 - 71.43 | 100 - 65.91 | 56 - 94.27|
| Mem Usage (MB-%)| 11.6 - 34.33 | 2.2 - 28.57 | 43.1 - 21.59 | 14.9 - 83.63|
## [944. Delete Columns to Make Sorted.][944]
I cannot see the best solution here. I'm definitely sure
mine isn't the one.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 90.91 | 8 - 100.00 | 88 - 82.50 | 112 - 81.53 |
| Mem Usage (MB-%)| 8.3 - 40.91 | 2.3 - 100.00 | 42.5 - 76.67 | 14.4 - 84.557|
## [961. N repeated elements in size 2N Array][961]
Two ways to go about this:
1. Build a set with 'seen' values. Break the moment we find a value we have already seen. Rust and Python use this.
2. Iterate through the array and check triplets. Since we know
that N-1 of the elements are the same, we're bound to bump into a triplet with two elements being equal. Edge case is 2N = 4, were the elements are in the beginning and end respectively.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 100.00 | 0 - 100.00 | 80 - 88.70 | 188 - 93.66%|
| Mem Usage (MB-%)| 7.2 - 73.91 | 2.1 - 100.00 | 41.9 - 76.92 | 15.3 - 65.97|
## [965. Univalued Binary Tree][965]
Traverse tree and bail when we find value that doesn't match root value. Iterative
this way we can bail fast.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 93.95 | 28 - 83.02%|
| Mem Usage (MB-%)| 6 - 45.00 | 2 - 100.00 | 39 - 44.59 | 14 - 95.67|
## [976. Largest perimeter triangle.][976]
Becomes easy when you find out about the
[triangle inequalities](https://en.wikipedia.org/wiki/Triangle_inequality) for when
a triangle with non zero (only `>` in the inequalities) can be formed.
Then, we sort our input array by size and start with maximum sides available. When the
inequality holds, we return the perimeter.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 28 - 100.00 | 4 - 100.00 | 96 - 98.61 | 172 - 98.82 |
| Mem Usage (MB-%)| 8 - 13.64 | 2.1 - 75.00 | 42.3 - 16.67 | 15.6 - 21.68 |
## [977. Squares of a Sorted Array][977]
Though a more convoluted way were we don't sort is
possible, I don't think it's of much worth.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 100 - 96.55 | 8 - 98.46 | 116 - 84.11 | 184 - 99.84 |
| Mem Usage (MB-%)| 20.3 - 33.10 | 2.2 - 78.46 | 45.4 - 39.09 | 16.1 - 34.75 |
## [985. Sum of even numbers after queries.][985]
Main trick here is to notice that, after finding the initial total sum of all even
values in the array, we have all information needed during each iteration to
update the sum and keep it up to date.
Whatever the case, if after updating a specific index we find that the new value is
even, we need to add it to the sum of evens. Then, if our value before updating was
even, we need to now remove it from the sum. It's like a switch of those two values.
In cases where the old/new value is odd, we can just ignore it. (since it didn't
contribute in any way to our sum).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 148 - 100.00 | 12 - 100.00 | 112 - 99.16 | 484 - 98.95 |
| Mem Usage (MB-%)| 21.6 - 56.52 | 2.7 - 100.00 | 47.1 - 24.37 | 19 - 97.14 |
## [989. Add to array from integer.][989]
Generalized version of One Plus/Plus One problem. We start from the end of the array
and start adding the digits in `K` from the right. We need to handle two cases, as
with the case of Plus One.
If the sum of the digit with the value in the array is
larger than nine, we add that sum `mod 10` to the array and increment `K` by one (`K`
has been reduced `/ 10` by this point).
If not (the sum is `< 9`) we just continue to the next digit in `K`.
If `K` reached `0` or we reach the start of the array, we stop. We then need to check
if `K` is zero or more values need to be added. If so, we create a temporary array
out of the digits and prepend it (reversed!) to `A`.
We then return either the new array or the original array `A`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 84 - 98.28 | 8 - 100.00 | 116 - 92.73 | 248 - 99.67 |
| Mem Usage (MB-%)| 15.1 - 100.00 | 2.2 - 100.00 | 43.3 - 86.36 | 14.9 - 93.23 |
## [997. Find the town judge.][997]
Use a list/array to keep track of who respects who (if we find someone as respecting someone else, we
mark its entry with `-1` to ignore it thereafter). It then boils down to traversing the list and checking
if an entry has `n-1` people respecting them.
Runtime/memory complexity is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 136 - 97.83 | 12 - 100.00 | 96 - 99.53 | 708 - 95.84 |
| Mem Usage (MB-%)| 16.5 - 14.13 | 2.8 - 37.50 | 45.9 - 92.09 | 19 - 62.15 |
## [999. Available Captures for Rook.][999]
Find the position of the rook and then just check it's row and column. (But from
the rooks position outwards.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 87.42 | 20 - 99.12 |
| Mem Usage (MB-%)| 5.6 - 76.47 | 2.1 - 75.00 | 37.6 - 99.34 | 14.1 - 83.77 |
## [1002. Find common characters][1002]
**TODO: See again.**
Build counts and go through values of string and find common characters.
This probably isn't the best way to go about it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 8.33 | 4 - 45.45 | 88 - 81.43 | 36 - 97.01 |
| Mem Usage (MB-%)| 12.1 - 5.56 | 2 - 100.00 | 42.7 - 55.97 | 14 - 98.23 |
## [1008. Construct Binary Search Tree from preorder traversal.][1008]
Key insight is that left subtree is composed of values smaller than the parent
value and right of values larger than parent value.
During each recursive call, we pass bounds to the function indicating the values that
should be considered for left/right subtrees. Initially, these denote the full array.
Then we scan the array and find the maximum value that is smaller than our parent value
, this is the bound for the values of the left subtree. The rest denote the values
for the right subtree.
With a base case of the bounds matching, we can recursively build our tree.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 99.36 | 32 - 92.64 |
| Mem Usage (MB-%)| 8.1 - 60.78 | 2.1 - 66.67 | 40.3 - 19.23 | 14.3 - 46.43 |
## [1009. Complement of a base-10 integer.][1009]
[Same as problem.][476]
## [1021. Remove Outermost Parentheses][1021]
Go through string and build resulting string with the help of a
depth counter tracking how many parentheses we've seen.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 90.17 | 28 - 98.78 |
| Mem Usage (MB-%)| 6 - 5.77 | 2.2 - 100.00 | 41 - 6.21 | 14.3 - 100.00 |
## [1022. Sum of root to leaf binary numbers][1022]
As we DFS our way to the bottom, we alter each node's value
to add what we've already seen (parent's value). This can
be done directly by using:
```python
node.val = (parent.val << 1) | node.val
```
When we reach a leaf, we add the value to an array which we
sum when we've visited every node.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 85.42 | 0 - 100.00 | 84 - 80.08 | 28 - 97.19 |
| Mem Usage (MB-%)| 7.2 - 14.58 | 2.1 - 100.00 | 41.2 - 16.26 | 14.8 - 13.95 |
## [1025. Divisor Game][1025]
Apparently, optimal move is dividing by one. Return answer based on if N
is odd or even.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 87.19 | 24 - 92.61 |
| Mem Usage (MB-%)| 5.4 - 98.95 | 2 - 92.00 | 38.4 - 57.93 | 14.1 - 78.14 |
## [1030. Matrix Cells in Distance Order.][1030]
The main gist is that we create a bunch of distances in ascending order, this can
be done by taking two ranges that are based on the following comparisons:
```python
limR = max(R-r0, r0) + 1
limC = max(C-c0, c0) + 1
```
and then combining them. By taking those ranges and not a combination of the
full range from `[0-R)` and `[0-C)` we eliminate many iterations that we
*know* will fall outside of the acceptable bounds. (I.e if `R` is `50` and
`r0` is `25` it doesn't make any sense to check for distance values `> 25`.)
Then we loop through them and try and apply them to the
initial point. If we fall in acceptable ranges `[0, R)` and `[0, C)` we can
keep it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 104 - 100.00 | 12 - 100.00 | 140 - 83.33 | 144 - 98.49 |
| Mem Usage (MB-%)| 23 - 12.50 | 2.7 - 75.00 | 47.5 - 40.00 | 16.2 - 41.04 |
## [1037. Valid Boomerang.][1037]
Check that the slopes for the two points are different (`y2 - y1 / x2 - x1`). Need
to check for edge cases first though (`x2 - x1 != 0`) and filter easy cases where
we know that they will be on the same line (two points being equal, all x's or all y's
for the points match.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.55 | 24 - 96.82 |
| Mem Usage (MB-%)| 5.6 - 62.96 | 2 - 85.71 | 38.8 - 39.66 | 14.3 - 14.43 |
## [1038. Binary Search Tree to greater sum tree.][1038]
This is a depth first inorder traversal. We do need to go right though. Care is
needed when traversing the left subtree; the accumulated value needs to be equal
to the value accumulated after visiting `left` and equal to the parent value if
the left subtree doesn't exist.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 92.61 | 24 - 96.32 |
| Mem Usage (MB-%)| 5.9 - 100.00 | 2 - 100.00 | 39.2 - 17.24 | 14.1 - 87.18 |
## [1046. Last stone weight.][1046]
This is the job for a heap that helps us retain sorted order after each pop/push.
`Python`, `JS` and `Rust` for which a heap or Priority Queue is available use the
heap approach.
For `C`, for which I will not write a heap now, uses the poor-mans solution of keeping
the invariant (sort-order) by continuously calling `qsort` after each loop.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 83.61 | 20 - 99.17 |
| Mem Usage (MB-%)| 5.5 - 100.00 | 2 - 83.33 | 39.2 - 75.82 | 14.2 - 29.54 |
## [1047. Remove All Adjacent Duplicate Strings][1047]
Use a stack to filter out adjacent elements.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 94.52 | 0 - 100.00 | 84 - 95.59 | 48 - 99.13 |
| Mem Usage (MB-%)| 8.1 - 9.59 | 2.2 - 60.00 | 47.1 - 42.68 | 14.7 - 21.89 |
## [1051. Height Checker][1051]
Create a sorted target array and go through both arrays
counting their differences.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.33 | 16 - 100.00 |
| Mem Usage (MB-%)| 5.7 - 70.92 | 2 - 40.00 | 38.4 - 81.16 | 14.1 - 69.05 |
## [1071. Greatest common divisor of strings.][1071]
Basically euclidean but for strings, mod is defined as slicing from the smaller string
from the larger string as many times as possible (using `starts_with/HasPrefix/strncmp`).
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 34 - 93.75 |
| Mem Usage (MB-%)| 5.8 - 90.00 | 2 - 59.26 | 2.3 - 83.67 | 13.8 - 75.49 |
## [1078. Occurences after Bigram.][1078]
Split the string and go through the words comparing. Use a while loop to skip over
chunks as needed. Similarly in `C` but with a significant ammount of boilerplate.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 98.41 | 20 - 98.86 |
| Mem Usage (MB-%)| 5.8 - 100.00 | 2 - 100.00 | 38.4 - 69.84 | 14.3 - 23.63 |
## [1103. Distribute candies to people.][1103]
Finally realized how n increases! The main trick here is to realize that
we can find the sum directly after `k` times filling the array.
For each `i` slot of our array, we increase each and every single time we
reach the end by `n`. After `k` steps, in the slot `i` for our result
array we will have:
```python
value_for_i * k + (n + 2n + 3n + 4n + ... + kn) ==
value_for_i * k + (1 + 2 + 3 + ... + k) * i ==
value_for_i * k + (k * ((k + 1) / 2)) * i
```
where `value_for_i` is `1, 2, 3, 4` depending on our index.
Using this (along with a small function to increase `k` until we reach close to
`candies`) we can find the next sum of the array values that is larger than
candies and stop. We then take the previous sum and fill in what's left.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 100.00 | 32 - 91.00 |
| Mem Usage (MB-%)| 5.9 - 79.17 | 2 - 83.33 | 38.3 - 88.75 | 14.1 - 97.00 |
## [1108. Defanging an IP Address][1108]
Instead of calling a replace function continuously, its
better to just go through the string and build a new string
with '.' replaced by '[.]'.
At least, that's my naive first stab at it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100 | 0 - 100| 72 - 79.62|32 - 45.31|
| Mem Usage (MB-%)| 5.8 - 14.33 | 0 - 100 | 38.6 - 8.88 | 14.1 - 99.95 |
## [1122. Relative sort array.][1122]
Create a dictionary holding the order in which elements should be
sorted. For elements not contained in the second array, we
use their value + 1000 to place in correct ascending order
(most dictionaries allow indexing that doesn't contain a key
to return a default value.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 88.20 | 0 - 100.00 | 72 - 97.44 |28 - 97.20|
| Mem Usage (MB-%)| 6.2 - 51.22 | 2 - 81.25 | 38.5 - 90.28 | 14.3 - 58.12|
## [1128. Number of equivalent domino pairs.][1128]
We first build a count of the pairs and then count their combinations.
The count can be built if a mapping supports lists as keys or, if we create a bijection
between the domino pairs and indices in an array. The first approach is used in Python
while the second in the rest (and my bijection isn't perfect but it suffices.)
After that we can use each count to count the number of combinations in for each pair.
The typical factorial equation is used to compute this:
```
(k + n - 1)! / k! * (n - 1)!
# which can be reduced, with k == 2 in our case to:
(n * (n - 1)) / 2
```
All in all, this is an `O(N)` time complexity algorithm with `O(1)` space (the # of pairs is
45 in total.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 32 - 100.00 | 4 - 100.00 | 92 - 93.75 | 212 - 100.00 |
| Mem Usage (MB-%)| 11.7 - 100.00 | 4.2 - 100.00 | 48.1 - 48.21 | 23.7 - 76.65 |
## [1137. N-th tribonacci number.][1137]
Not at all different from the normal fibonacci case. Iterative implementations,
we just keep track of three values during each iteration.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 97.31 | 24 - 93.78 |
| Mem Usage (MB-%)| 5.3 - 100.00 | 2 - 88.89 | 38.3 - 79.62 | 14.1 - 69.63 |
## [1146. Snapshot Array.][1146]
Keep an array of maps around. The maps associate, for each index of the array
a specific snapshot to value. Care is only needed when calling `get` to make sure
we return the right `snap_id` (one which equals the given `snap_id` or a snapshot
in the map with the largest value that's smaller than `snap_id`, i.e, the most recent
snapshot for that index). As a result, get is `O(number_of_snaps)`.
The results for Javascript, that uses the same logic, are.. interesting.
TODO: See again, also, add `C` when you're up for dealing with `uthash`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 48 - 100.00 | 388 - 35.90 | 376 - 95.81 |
| Mem Usage (MB-%)| | 26.4 - 25.00 | | 54.1 - 10.26 |
## [1160. Find Words That can be Formed by Characters.][1160]
**TODO: See again.**
Similar to problem [Find common characters][1002]. I, again, can't seem to see
what is better than building a Counter. Maybe a reg-exp solution? Not sure, I'll
might need to check the Discussions if I can't think of anything else.
*Maybe*, seeing results for Javascript now, I was overreacting previously.
**Pending**: `C` version (Which I'm suspecting will re-affirm initial
suspicion).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 8 - 100.00 | 92 - 97.14 | 92 - 87.95 |
| Mem Usage (MB-%)| | 2.1 - 100.00 | 46.1 - 78.91 | 14.7 - 11.99 |
## [1170. Compare strings by frequency of the smallest character.][1170]
The basic trick we can use is a cache holding, for every string in `queries` seen
so far, the ammount of strings that have a value for `f` larger than the query.
When we find another query with the same frequency we can then check the cache without
needing to go through the `words` array to count the words again.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 92.31 | 0 - 100.00 | 96 - 91.96 | 64 - 92.86 |
| Mem Usage (MB-%)| 7.7 - 46.15 | 2.5 - 33.33 | 46.2 - 25.00 | 14.9 - 10.41 |
## [1184. Distance between Bus Stops.][1184]
This can be solved by finding the sums until the start, destination and end. The
reasoning behind this is harder without a piece of paper to draw it on, unfortunately
but I'll try.
The first difference we want to consider is that when going from start to dest in
the direction of dest. That is, if `start < dest`, we go right, if not, we go left.
This, hopefully clearly, is `|sum(start) - sum(dest)|`, we only want the intermediate
elements.
The second difference we want to find is the one going away from dest (initially, at
least). That is, if `start < dest` we go left, if not, we go right.
In this case, we have:
- if `start < dest`, the difference is `sum(end) + sum(start) - sum(dest)`
(we count everything from start until zero, and then we count from end until dest).
- if `start > dest`, it is flipped: `sum(end) - sum(start) + sum(dest)` using the
same reasoning as before.
It might now be clear that the second difference is `sum(end) - difference_one` where
difference one is the first difference.
Runtime complexity is necessarily `O(N)` since we need to build the sum until the end.
Space complexity is `O(1).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.00 | 40 - 94.38 |
| Mem Usage (MB-%)| 6.2 - 97.06 | 2.1 - 100.00 | 38.6 - 76.00 | 14.8 - 98.70 |
## [1185. Day of the week.][1185]
These all use Pythons `datetime.date.weekday` algorithm for finding
the day name.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 89.82 | 24 - 93.06 |
| Mem Usage (MB-%)| 5.5 - 92.86 | 1.9 - 50.00 | 38.3 - 93.81 | 14.2 - 41.01 |
## [1189. Maximum number of balloons.][1189]
Create a counter out of the characters in the string and then continously
reduce count of characters composing 'balloon' until we can't anymore (or
we find that a character doesn't exist in the counter).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 96.63 | 28 - 91.08 |
| Mem Usage (MB-%)| 5.9 - 27.12 | 2.1 - 75.00 | 40.6 - 44.17 | 14.4 - 13.09 |
## [1200. Minimum Absolute Difference.][1200]
Sort the array and then go through it and add the minimum distances to our
result (minimum difference might change while iterating, in which case we
dump all previous contents of result array and start fresh.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 84 - 100.00 | 16 - 80.00 | 160 - 88.54 | 296 - 99.61 |
| Mem Usage (MB-%)| 16.3 - 88.24 | 3.1 - 100.00 | 48.7 - 100.00 | 28.1 - 33.97|
## [1207. Unique number of occurences.][1207]
Create counts for each value. If, by using a Set, any values
are filtered, we've got duplicates.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 94.09 | 28 - 94.97 |
| Mem Usage (MB-%)| 6.5 - 23.94 | 2.1 - 80.00 | 38.6 - 80.38 | 14.2 - 83.42|
## [1217. Minimum cost to move chips to the same position.][1217]
Count evens and odds. Answer is the minimum between those two
since all values of `max{evens, odds}` can be moved to one pile
via 2-steps (with zero cost) leaving the `min{evens, odds}` as
the cost we need to count (note that no matter their distance.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 74.31 | 24 - 95.45 |
| Mem Usage (MB-%)| 5.8 - 63.86 | 2 - 94.23 | 37.8 - 96.26 | 14.1 - 78.87|
## [1221. Split a string in balanced strings.][1221]
Couple of variables needed to track what character a new sequence of balanced pairs
starts with along with how many occurences of that character we've seen. After that we
just decrement when we see a different char and increment when we see the same char.
All in all, `O(N)` runtime and `O(1)` memory.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 94.44 | 24 - 95.29 |
| Mem Usage (MB-%)| 5.5 - 63.69 | 2 - 95.45 | 38 - 96.47 | 14.3 - 10.74 |
## [1237. Find positive integer solution for equation.][1237]
Use the fact that the function is increasing to bail early in second
loop.
Note: Though this is wrong, output values only range in [1, 100] range. This
basically allows us to skip many iterations by only going through that
range. This is wrong and I should feel wrong.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 98.17 | 28 - 97.04 |
| Mem Usage (MB-%)| 6 - 96.15 | 2 - 100.00 | 38.8 - 71.34 | 14 - 83.57|
## [1252. Cells with Odd Values in a Matrix][1252]
**TODO: Describe algorithm better.**
Note: Peeked at hints for this problem.
The tldr; of the algorithm boils down to this. If both `m` and `n`
are `> 2`; the sum of all odd elements in the matrix can be
calculated by solely knowing the number of odd rows and odd
columns.
Since we toggle between odd and even numbers we can keep track
of which rows and which columns are left in an odd state and
then sum of the cells according to a relatively (after some hours
and many printed arrays later) simple formula:
```python
(m - num_odd_cols) * num_odd_rows + (n - num_odd_rows) * num_odd_cols
```
The minor edge cases `n or m <= 2` can be handled separately.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.98 | 20 - 100.00|
| Mem Usage (MB-%)| 6 - 98.61% | 2 - 100.00 | 39.1 - 5.66 |14 - 7.53|
## [1260. Shift 2d grid.][1260]
By taking the value of `k mod cols` we can easily find how many rows/columns
we need to skip and bail early on certain cases.
With Python and Rust, we can just utilize a deque's `rotate` method after
flattening the array.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 60 - 100.00 | 8 - 100.00 | 104 - 84.38 | 144 - 97.29 |
| Mem Usage (MB-%)| 14.3 - 89.19 | 2.3 - 33.33 | 46.1 - 10.42 | 14.4 - 81.01 |
## [1266. Minimum Time Visiting All Points][1266]
Move as much as you can diagonally and then move horizontically or
vertically according to the case. `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 98.64 | 0 - 100.00 | 76 - 89.95 | 52 - 96.32 |
| Mem Usage (MB-%)| 6.4 - 13.61 | 2.1 - 100.00 | 40 - 5.60 | 14.1 - 100.00|
## [1275. Find winner on a tic tac toe game.][1275]
Mostly straighforward, fill a board up with the value and check if we have a
winner.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.18 | 28 - 90.26 |
| Mem Usage (MB-%)| 5.6 - 75.00 | 2 - 100.00 | 38.6 - 84.94 | 14.2 - 90.41 |
## [1281. Subtract the Product and Sum of Digits of an Integer][1281]
Continuously reduce `n` by `// 10` and add/multiply its modulo
`10` (the rightmost digit).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 86.84 | 20 - 98.44 |
| Mem Usage (MB-%)| 5.7 - 14.29 | 2 - 100.00 | 38.5 - 20.85 | 14.1 - 99.96 |
## [1282. Group the people given the group size they belong to.][1282]
Use a map to hold the groups as we build them. Once a group is filled up,
add it to the result and reset the group in the map.
Don't forget to add any pending groups after iterating through `groupSizes`.
Runtime complexity is `O(N)` where `N` is the size of `groupSizes`, space
complexity is, I believe, similarly `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 24 - 100.00 | 0 - 100.00 | 84 - 100.00 | 72 - 88.83 |
| Mem Usage (MB-%)| 9.7 - 80.00 | 2.1 - 72.00 | 43.9 - 84.29 | 14.2 - 98.32 |
## [1287. Element appearing more than 25% in sorted array.][1287]
Calculate the threshold 25% corresponds to and go through the array counting lengths
of equal items. If at any point during our counting we exceed the threshold, we return
immediately.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 100.00 | 0 - 100.00 | 76 - 92.64 | 68 - 99.84 |
| Mem Usage (MB-%)| 6.4 - 65.31 | 2.1 - 100.00 | 39.5 - 82.68 | 15.42 - 35.05 |
## [1290. Convert Binary Number in a Linked List to Integer][1290]
Traverse the list adding the sum as we go. `O(N)` necessarily.
`O(1)` space when adding the sum as we go.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00% | 0 - 100.00% | 76 - 50.45% | 20 - 98.93%|
| Mem Usage (MB-%)| 5.6 | 1.9 - 100.00% | 38.6 - 25.42% | 14.2 - 100.00%|
## [1295. Find Numbers with Even Number of Digits][1295]
See the Python (1295.py) for the more generic approach in which
we continuously divide by ten and use a toggle to check whether
our divisions are odd or even until reaching zero (ok, until
reaching [10, 100] as a minor opt).
Hardcoding the ranges, though specific, is fastest.\*
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 | 36 - 99.95 |
| Mem Usage (MB-%)| 6.3 - 99.58 | 2 - 8.96 | 38.2 - 6.75 | 14.3 100.00 |
\* I've used the fastest times for the table.
## [1299. Replace Elements with Greatest Element on Right Side][1299]
Go backwards, continously updating the maximum value seen as we go through.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 52 - 90.45 | 4 - 96.67 | 88 - 96.36 | 104 - 99.74 |
| Mem Usage (MB-%)| 12.7 - 48.64 | 2.2 - 100.00 | 42.1 - 5.56 | 15.3 - 6.78 |
## [1302. Deepest leaves sum.][1302]
Traverse tree keeping track of current depth, max depth and the sum. If we encounter
a leaf we check its depth.
If it's larger than the current depth, reset the sum to the
leaf value and keep track of new maximum depth. Else, add leaf value to sum.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 97.01 | 8 - 100.00 | 96 - 97.42 | 84 - 94.32 |
| Mem Usage (MB-%)| 15.8 - 83.58 | 3.1 - 100.00 | 48.7 - 71.94 | 17.8 - 70.79 |
## [1304. Find N Unique Integers Sum up to Zero][1304]
Fill the array up from the range `[i, (n/2) + 1)` with `i` and
`-i` values. If the array is odd `n & 1 == 1`, add a `0` to
complete it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 94.87 | 0 - 100.00 | 76 - 88.60 | 24 - 97.78 |
| Mem Usage (MB-%)| 6.8 - 27.35 | 2 - 8.33 | 39.1 - 9.00 | 14.3 - 100.00 |
## [1309. Decrypt String from Alphabet to Integer Mapping][1309]
In both cases that follow, we go from one character or set of
characters to another. Since we can work with characters using
their code points, we can calculate the correct mapping just by
adding the appropriate numbers. The offset from going from `'1'`
to `'a'` is `48`. That is, the code point for `'1'` (`49`) plus
the offset of `48` leads to `97` which is the code point for `a`.
As such, whatever the case, we know we need to cover the offset
of `48` by adding it.
First we through the characters in the string and check two cases:
1. The second character from the one we're looking at is a `'#'`.
2. It isn't. In this simpler case, we just convert into the
appropriate character using their code point value.
In the first case, we then look at the following character
from the one we're currently at. If it is a `1` we know we're
looking at the set of the *ten* next characters in the mapping.
As such, we acount for these by adding `10` to the offset of `48`
before adding the value for the code point.
Similarly, if the following character is a `2`, we are looking
at the final remaining 7 characters. As such, we add `20` to
the offset to get to the correct output character.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 88.44 | 20 - 99.29 |
| Mem Usage (MB-%)| 5.8 - 9.46 | 2 - 100.00 | 38.9 - 5.19 | 14.2 - 100.00 |
## [1313. Decompress Run-Length encoded List][1313]
**TODO: Improve**
I feel I haven't found best solution here, need to look into
it, especially the mem usage in Python bothers me.
(Yea, C code is also mediocre, there's
definitely something better here.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 32 - 81.07 | 4 - 100.00 | 92 - 85.55 | 64 - 89.35|
| Mem Usage (MB-%)| 10.4 | 2.1 - 100.00 | 42.3 - 5.85 | 14.5 - 13.10|
## [1315. Sum of nodes with even valued grandparent.][1315]
Traverse the tree normally, hold around the value of the parent of the node
currently traversing on and if that node's children exist and the parent value
is even, add them to the sum. Recurse.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 93.48 | 4 - 100.00 | 96 - 99.00 | 92 - 94.27 |
| Mem Usage (MB-%)| 15.7 - 97.83 | 3 - 100.00 | 48.8 - 62.50 | 17.7 - 93.81 |
## [1317. Convert integer to the sum of the two non-zero integers.][1317]
Start from the largest, smallest pair possible `(n-1, 1)` and either increase the
smallest if their sum is smaller than `n` or decrease largest if sum larger than `n`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 92.63 | 24 - 94.60 |
| Mem Usage (MB-%)| 5.5 - 50.00 | 2 - 100.00 | 38.4 - 96.84 | 14.2 - 66.61 |
## [1323. Maximum 69 Number][1323]
**TODO: Improve. I don't think this is best solution.**
After finding the largest 10 multiple `d` of 9 that has the same
number of digits as the number, we continuously divide until
we can't anymore (signifying a `6`, consider `669 // 900`).
When we reach that point we simply add `d // 3` to our original
number (again, consider `669 // 900`, we need `900 // 3 == 300`
to reach our desired result `969`).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 47.88 | 20 - 98.80 |
| Mem Usage (MB-%)| 5.4 - 29.69 | 1.9 - 100.00 | 38.8 - 32.54 | 14.1 - 99.85 |
## [1329. Sort the matrix diagonally.][1329]
The trickiest part is finding how to nicely traverse the diagonals of the matrix.
After doing that, we can collect each diagonal in a separate vector, sort it and
then place it back into the matrix.
I've implemented diagonal traversing via while loops. Starting from the leftmost
bottom position, we decrease the row index and (after the row index reaches zero)
icrease the column index. The inner while loops simply traverse the diagonal (each
index is increased by one until we reach the end for either one).
Can't really guess much on the complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 91.92 | 0 - 100.00 | 84 - 99.46 | 76 - 96.70 |
| Mem Usage (MB-%)| 8.5 - 90.91 | 2.2 - 90.38 | 41 - 92.90 | 14.6 - 52.14 |
## [1332. Remove palindromic subsequences.][1332]
**Big** note to self: *Pay attention to the description*. It asks for
**subsequences**, not **substrings**.
Number of removals can be found quickly since we have subsequences:
- Empty string => 0 ops.
- A full palindrome => 1 op.
- Else: create set, it's size will be number of ops. (i.e first op remove all a's
and on second op remove all b's.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 96.10 | 20 - 98.28 |
| Mem Usage (MB-%)| 5.5 - 100.00 | 2 - 100.00 | 39.8 - 6.49 | 14 - 95.81 |
## [1337. The k-weakest rows in a matrix][1337]
Sum each row and enumerate it, sort and then grab `k` first.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 100.00 | 0 - 100.00 | 76 - 95.42 | 96 - 98.53 |
| Mem Usage (MB-%)| 6.8 - 55.56 | 2.1 - 50.00 | 40 - 81.67 | 14.5 - 57.69 |
## [1342. Number of Steps to Reduce a Number to Zero][1342]
Xor with 1 if even (to drop the 1) and right shift by 1 to divide by two. Count.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 78.28 | 28 - 78.69 |
| Mem Usage (MB-%)| 5.5 - 27.87 | 2 - 100.00 | 38.8 - 29.88 | 14.1 - 99.98 |
## [1346. Check if n and its double exist.][1346]
This follows a pretty common trick whereby you add items into a set and check new
number against those in the set to see if the condition you're looking for holds.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 72 - 90.38 | 44 - 96.77 |
| Mem Usage (MB-%)| - | 2.4 - 5.88 | 40.6 - 23.06 | 14.3 - 51.18 |
## [1351. Count Negative Numbers in a Sorted Matrix][1351]
**TODO: Describe logic. Overview of steps as seen in 1351.py**
The main idea is by going to the edges of the matrix (top-right
corner and bottom-left) we can bulk count negative numbers by
using the fact that the matrix is row and column sorted.
If we find negative cells, we count everything after it. After
doing that, we adjust our points to move around the matrix
looking for the next negative cell.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 97.86 | 0 - 100.00 | 76 - 85.01 | 104 - 99.49 |
| Mem Usage (MB-%)| 7.1 - 77.14 | 2.3 - 100.00 | 39.9 - 7.13 | 14.9 - 99.62 |
## [1356. Sort integers by the number of 1 bits][1356]
Sort initially to get them in correct order. Then sort by count. The second sort
*must* be stable. In `C` this is done by creating a new array holding required
information while in `Javascript` we use `lodash`. `Rust` and `Python` offer
stable sorts.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 89.87 | 0 - 100.00 | 96 - 82.93 | 56 - 97.11 |
| Mem Usage (MB-%)| 8.3 - 16.46 | 2 - 92.59 | 42.3 - 40.85 | 14.1 - 92.41 |
## [1365. How Many Numbers Are Smaller Than the Current Number][1365]
By sorting the array we can find the # of elements faster in
one pass through the array since we know that all consequent
elements will be larger (or smaller, depending on sort type).
Using a Map we can then keep track of the occurences of elements already seen and set that as the value for an element.
Finally we can go through the nums array and for every element
get the occurence for it and place it in the final resulting
array.
We use quite the extra space but avoid O(N^2) this way
(and instead result in O(nlogn)).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 77.53 | 0 - 100.00 | 92 - 76.99 |44 - 99.38|
| Mem Usage (MB-%)| 7.4 - 64.10 | 2.2 - 100.00 | 40.9 - 9.53 |14 - 100.00|
## [1370. Increasing Decreasing String.][1370]
Create three supporting structures:
- A sorted list of distinct characters.
- A reverse sorted list of distinct characters.
- A count of the # of occurences for each character.
Then, by toggling between the sorted and the reverse sorted list,
we decrease the counts of the counts mapping. This is done until
we reach 1. We then break and add all remaining elements. (Of
course, we could add them and *then* break but meh.)
**TODO for C: Create a generic map that's correct and can be re-used. Add it
to folder so I can re-use it when needed.**
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 13.46 | 4 - 90.00 | 100 - 80.77 | 44 - 100 |
| Mem Usage (MB-%)| 8.8 - 100.00 | 2.3 - 100.00 | 45.8 - 8.24 | 14.1 - 100.00|
## [1374. Generate a String With Characters That Have Odd Counts][1374]
Basically, two cases:
- `n` is odd: Repeat one character `n-2` times and add two others.
- `n` is even: Repeat one character `n-1` times and add one other.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 87.27 | 20 - 98.14 |
| Mem Usage (MB-%)| 6.2 - 74.04 | 2.2 - 100.00 | 39 - 19.09 | 14.3 - 100.00 |
## [1379. Find a corresponding node of a binary tree in a clone of that tree.][1379]
Just traverse the cloned tree looking for a node with a value equal to the value of
the target node.
Note on Follow up: For this case, we can simply build a path in the original tree
while looking for target and then follow that path in cloned to find its reference.
Note: C and Rust versions aren't available.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| N/A | N/A | 296 - 94.16 | 592 - 97.39 |
| Mem Usage (MB-%)| N/A | N/A | 59.4 - 70.94 | 23.9 - 96.23 |
## [1380. Lucky numbers in a Matrix][1380]
Build a set to hold minimum values of each row and a
list/array to hold maximum values for each column.
After iterating through each row and filling the previous
structures us, return any elements in the array/list that
are also present in the set.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 16 - 100.00 | 0 - 100.00 | 76 - 96.19 | 116 - 95.12 |
| Mem Usage (MB-%)| 7 - 17.65 | 2.1 - 100.00 | 40.7 - 87.20 | 14.3 - 49.67 |
## [1381. Design a stack with increment operation.][1381]
The implementation is pretty straight-forward for all cases. I do feel like I'm
missing something (due to the Python timings). All operations are `O(1)` (`C` malloc
initialization, not sure what the complexity is there, though), except for
`increment` which is `O(k)`.
TODO: Check again.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 32 - 91.07 | 4 - 100.00 | 124 - 88.82 | 112 - 76.56 |
| Mem Usage (MB-%)| 13.7 - 85.71 | 2.9 - 50.00 | 46.3 - 50.31 | 14.7 - 98.36 |
## [1385. Find the distance between two arrays.][1385]
Go through fist array and for each value make top and bottom bounds for
values of second array to satisfy in order for `| arr1[i] - arr2[j] <= d |`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 85.71 | 0 - 100.00 | 88 - 87.00 | 76 - 89.36 |
| Mem Usage (MB-%)| 6.1 - 64.29 | 2.1 - 100.00 | 39.2 - 99.00 | 14.1 - 98.84 |
## [1386. Sort integers by the power value.][1386]
TODO: Revisit. Seems like rust has a better approach (tabulation, I'd guess). Couldn't be
bothered with `C` because of the cache, could use array with max length `hi * 3 + 1` I'd guess.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 24 - 42.86 | 116 - 87.97 | 56 - 99.03 |
| Mem Usage (MB-%)| | 2 - 85.71 | 45.2 - 20.30 | 14.5 - 67.13 |
## [1389. Create target array in the given order.][1389]
O(N^2) is actually best option here, no secret fancy solutions. Continuously
insert in target position.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 91.95 | 28 - 91.36 |
| Mem Usage (MB-%)| 5.9 - 94.44 | 2.1 - 27.78 | 40.2 - 9.52 | 14 - 88.59 |
## [1394. Find lucky integer in Array.][1394]
Build a counter and return the largest pair for which `key == value`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 97.56 | 0 - 100.00 | 68 - 100.00 | 48 - 96.34 |
| Mem Usage (MB-%)| 6.4 - 31.71 | 2 - 33.33 | 40 - 50.62 | 14.1 - 97.03 |
## [1399. Count largest group.][1399]
Build a Counter, sort it and find the length of the group with the largest
values.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 55.56 | 4 - 100.00 | 84 - 93.48 | 80 - 90.56 |
| Mem Usage (MB-%)| 6 - 11.11 | 2 - 100.00 | 40.9 - 55.43 | 14.1 - 86.40 |
## [1403. Minimum Sequence in non increasing order,][1403]
Get the sum of the sequence and then sort it (ideally
descending). Initialize a value for counting the sum of
the subsequence seen so far. Then for every value in the
sequence:
1. sum the current value to the subsequence sum.
2. subtrack the current value from the sum of the sequence.
3. if the sum from 1 is strictly larger than that of 2, we are done and we return the slice until that point. Else
we continue through the values.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 96.00 | 0 - 100.00 | 84 - 92.78 | 52 - 97.26 |
| Mem Usage (MB-%)| 6.5 - 88.00 | 2 - 75.00 | 39.7 - 91.67 | 14.2 - 63.63 |
## [1408 String matching in an Arrat,][1408]
Sort array from smallest string to largest. Go through it comparing smaller with
larger strings. Bail when you find a match.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 91.22 | 28 - 97.24 |
| Mem Usage (MB-%)| 6 - 100.00 | 2 - 100.00 | 38.2 - 99.62 | 14.2 - 67.69 |
## [1409. Queries on a permutation with key.][1409]
Don't think there's much of a trick here, we go along with what the problem describes,
finding the index and then swapping the position of the value queried.
Runtime complexity should be O(nm) where n is the size of the queries.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 73.33 | 0 - 100.00 | 76 - 99.25 | 48 - 93.41 |
| Mem Usage (MB-%)| 6.1 - 98.33 | 2 - 81.82 | 40.1 - 67.91 | 14.4 - 84.89 |
## [1413. Minimum Value to get positive step by step sum.][1413]
Start with an initial minimum guess of `1` and go through the values of the
array. When we fall under `1` after summing values, we adjust the minimum value
in order to not be under `1`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 82.35 | 24 - 96.60 |
| Mem Usage (MB-%)| 5.8 - 100.00 | 2 - 100.00 | 37.9 - 97.39 | 14 - 96.50|
## [1414. Find the minimum number of fibonacci numbers whose sum is K.][1414]
Initially, I believed this was a dp-problem. Solving it like that (though possible)
was actually not the best way. Apparently, greedily subtracting largest possible
fibonacci number from `k` continuously yields the correct result.
Though I am sure this results in a possible answer (we simply subtract a fibonacci number from a
sum of n fibonacci numbers) I am still unsure why it is optimal.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 100.00 | 28 - 95.34 |
| Mem Usage (MB-%)| 5.4 - 72.73 | 2 - 100.00 | 38.7 - 100.00 | 14.3 - 11.66 |
## [1417. Reformat the string.][1417]
Build a list of characters and numbers. If their lengths don't differ by a maximum
of `1`, we cannot reformat the string as needed. If they do, we just interleave
them (taking care on which we interleave first/second etc).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 98.72 | 28 - 100.00 |
| Mem Usage (MB-%)| 6.1 - 42.31 | 2.1 - 60.00 | 42.5 - 89.10 | 14.1 - 86.07 |
## [1422. Maximum Score After splitting a string.][1422]
Start from position 1 and count all ones to the right. Add one to the count if
position `0` has the character `'0'`.
Go through the strings with the count as the current max. If a `'1'` is seen, we
subtract one from the count, if `'0'` is seen, we add one (this is because left
hand string counts zeroes and right hand string counts ones.
After each iteration we check if the new count is larger than the max, if so, keep
it. After iterating through the string, we will have kept the largest value for max.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 97.73 | 28 - 92.27 |
| Mem Usage (MB-%)| 5.6 - 94.74 | 2 - 100.00 | 38.7 - 86.36 | 14 - 91.17 |
## [1431. Kids with greatest number of candies][1431]
Find max, compare values and return.
## [1436. Destination City.][1436]
Use two sets, sources and destinations, and return the result
of the difference between destinations and sources (We know
that it will contain only one result due to the constraints.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 95.41 | 0 - 100.00 | 80 - 78.81 | 44 - 98.44 |
| Mem Usage (MB-%)| 7.6 - 48.62 | 2.4 - 100.00 | 41.7 - 5.08 | 14.1 - 100.00|
## [1437. Check if all 1s are at least k places away.][1437]
Does an `O(N)` scan of the array. Looks for ones and counts the difference between consecutive ones,
pretty straightforward.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 60 - 88.73 | 8 - 66.67 | 84 - 91.49 | 516 - 100.00 |
| Mem Usage (MB-%)| 11.1 - 80.23 | 2.5 - 100.00 | 46.5 - 73.05 | 16.7 - 96.80 |
## [1441. Build an Array with stack operations][1441]
Go through values of the target and build the operations. We build the
sequence of "Push" and "Pop"s for missing elements by finding the difference
of the current value of the target with its previous value.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 92.27 | 24 - 96.01 |
| Mem Usage (MB-%)| 6.2 - 70.00 | 2 - 100.00 | 38.5 - 82.73 | 14.1 - 84.41 |
## [1446. Consecutive Characters,][1446]
Go through characters and count.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 96.99 | 24 - 99.90 |
| Mem Usage (MB-%)| 5.8 - 98.72 | 2 - 98.78 | 39.3 - 64.05 | 14.3 - 12.21 |
## [1450. Number of Students Doing Homework at a Given Time][1450]
Go through first vector (start times) and for every start time
that is smaller equal to the queryTime, check the second vector
(end times) to see if we fall in range.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 86.58 | 28 - 97.73 |
| Mem Usage (MB-%)| 5.8 - 80.17 | 2.2 - 100.00 | 38.3 - 29.39 | 14.2 - 100.00|
## [1455. Check if a word occurs as a prefix of any word in a sentence.][1455]
Straight-forward approach. Split and use string functions for `Rust`, `Python`
and `Javascript`. Compare characters in `C`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 95.51 | 20 - 98.95 |
| Mem Usage (MB-%)| 5.5 - 100.00 | 2 - 100.00 | 38.4 - 42.31 | 14.1 - 41.63|
## [1460. Make Two Arrays Equal by Reversing Sub-Arrays][1460]
Sort and compare element-wise.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 97.78 | 0 - 100.00 | 72 - 99.53 | 64 - 99.47 |
| Mem Usage (MB-%)| 6.6 - 26.67 | 2.1 - 50.00 | 40.6 - 46.34 | 14.3 - 79.93 |
## [1464. Maximum Product of Two Elements in an Array][1464]
Though sorting the array and grabbing the last two elements seems
like the best way to do this, it's better even to find the max
value, then remove it from the array/vector and then call max
again.
Sorting alone is `O(nlogn)`.
Max, remove(1) and max is `O(N) + O(N) + O(N) => O(N)`.
(1): remove can also be done in O(1) if we have the index of the
first maximum. We don't remove it but instead replace its value
with a minimum value thereby removing it from consideration when
we find the next max.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 92.82 | 0 - 100.00 | 76 - 84.17 | 40 - 98.94% |
| Mem Usage (MB-%)| 6 - 99.52 | 2.3 - 100.00 | 38.6 - 6.57 | 14.2 - 100.00%|
## [1470. Shuffle the Array][1470]
**TODO: Come back to this.**
`C`: Go through the array and select appropriate place to grab
from based on if the index if odd or even.
`Python`: `chain` and `zip` take care of this quickly. Since
I'm more familiar with these, I opted for the one-liner.
`JavaScript`: Again, a more functional approach whereby I first
grab the half of the lift (until `n`) and then, using `flatMap`
I flatten and then return `[index, index+n]` sublists.
`Rust`: Tricky doing this in a functional way. Need to come
back to this after further practice.
## [1475. Final Prices With a Special Discount in a Shop][1475]
Brute force this. For each element of the array loop through
the rest of the array looking for the min element.
Can, at least, re-use the array we're given.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 98.68 | 0 - 100.00 | 80 - 91.33 | 44 - 97.07 |
| Mem Usage (MB-%)| 6.6 - 56.58 | 2.2 - 100.00 | 40.2 - 5.61 |14.3 - 100.00|
## [1476. Subrectangle Queries.][1476]
Can't see a non-O(N\*M) way to implement `update_subrectangle`, so, we brute force
it there. `get_value` is an easy `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 32 - 100.00 | 8 - 100.00 | 100 - 97.37 | 156 - 86.78 |
| Mem Usage (MB-%)| 10.2 - 85.71 | 3.1 - 93.75 | 45.2 - 11.18 | 15.8 - 99.72 |
## [1480. Running Sum of 1d Array][1480]
Basically add as we go through the array.
Also, basically the same as `itertools.accumulate`.
## [1486. XOR Operation in an Array][1486]
Just go through the 'array' (we don't need to build it obviously)
and reduce using xor.
In Python, we take advantage of `functools.reduce` being implemented in `C`, in the others we just loop through.
Again with some memory issues in `C` and `JS` though. I'm pretty
sure a XOR-trick for sequences should exist (appart from the
common `x ^ x = 0` and `x ^ 0 = x`; can't seem to find it though.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 86.39 | 28 - 78.99|
| Mem Usage (MB-%)| 5.7 - 22.45 | 2.2 - 100.00 | 38.2 - 5.14 |14.2 - 100.00|
## [1491. Average Salary excluding the min and max][1491]
Sort and then just sum and divide.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 89.32 | 24 - 94.65 |
| Mem Usage (MB-%)| 5.8 - 70.69 | 1.9 - 75.00 | 38.3 - 83.59 | 14.1 - 80.75 |
## [1496. Path Crossing.][1496]
Hold previous points visited in a set, after moving to a new point, check if it is
contained in the set.
`C` case: Quite frankly, not up for create a uthash structure with a compound key
today.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| N/A | 0 - 100.00 | 72 - 94.74 | 20 - 98.76 |
| Mem Usage (MB-%)| N/A | 2 - 100.00 | 38.5 - 74.34 | 14.4 - 60.40 |
## [1502. Can make arithmetic progression from sequence.][1502]
Sort the array and find the step by subtracting first two
elements. Go through the array and check that step is
present through-out.
Worse-case `O(NlogN)` due to sorting.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 89.04 | 0 - 100.00 | 76 - 88.50 | 32 - 95.03 |
| Mem Usage (MB-%)| 6.1 - 76.71 | 1.9 - 100.00 | 38.7 - 86.58 | 14.1 - 89.94 |
## [1507. Reformat date.][1507]
Split, and re-build. The main issue is taking care the starting `'0'` character needs
to be added on days `< 9` and months `< 9`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 91.22 | 24 - 94.80 |
| Mem Usage (MB-%)| 5.5 - 88.89 | 2 - 50.00 | 38.5 - 72.97 | 14.2 - 61.80 |
## [1512. Number of Good Pairs][1512]
Brute force leads to definite O(N^2). As such, we can first
sort the array (O(nlogn)) and then, instead of comparing
everything, notice that for a continuous sequence of equal
elements we have:
```python
[1, 1, 1, 1, 1, ...., 1, 2, 2, ...]
```
Starting from index `0`, as long as we keep on encountering the same number, we increase a counter. Then, when we reach the end (find
a number not equal to previous one), we notice that the number of
pairs thus far should be `n(n+1)/2` because the sequence of
comparisons grows as:
```
1 + 2 + 3 + ... + n
```
Repeat this until we reach the end.
We could always use some combination of group-by's, len to
move the explicit for loop in the backround.
## [1518. Water Bottles.][1518]
Continuously reduce `num_bottles` by dividing it by `num_exchange`, the quotentis added to the total number of drinks we can consume and `num_bottles` is
reassigned to `quotent + remainder` and the same steps are taken again until
the quotent becomes zero.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 89.06 | 24 - 93.26 |
| Mem Usage (MB-%)| 5.4 - 100.00 | 1.9 - 95.24 | 38.7 - 16.67 | 13.9 - 99.70 |
## [1523. Count Odd numbers in an interval range.][1523]
Can find odds by dividing the range by 2 and then adding one if at least one
of the bounds was odd. (Note: yes, you got this to work, but why +1?).
Space/time complexity is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.72 | 24 - 94.69|
| Mem Usage (MB-%)| 5.3 - 100.00 | 2 - 75.00 | 38.3 - 70.59 | 14.2 - 42.51 |
## [1528. Shuffle String][1528]
Create a new array and fill it in, basically.
`O(N)` time *and* `O(N)` space. I'm sure there should be a
way for inplace swaps.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 88.85 | 0 - 100.00 | 76 - 97.38 | 56 - 65.00|
| Mem Usage (MB-%)| 6.2 - 100.00 | 2 - 100.00 | 40.3 - 10.15 |14.1 - 99.95|
## [1534. Count Good Triplets][1534]
Not good. Can't think of a way to reduce the combinatorial
explosion by removing elements from the array that will never
satisfy the conditions.
I thought maybe some elements can be disqualified beforehand but, that doesn't seem to be the case (running the code with test data shows that all elements can appear). As such, I don't know how
to reduce the complexity.
**Update:** Remembered the 3-way set disjointness problem I had
encountered. We enter the third inner loop only if one of the
inequalities is satisfied (for which we need two inner loops to
check, see code). This way, we can skip many of the combinations
that don't match and reduce the overall explosion.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 100.00 | 0 - 100.00 | 76 - 97.33 | 288 - 98.17 |
| Mem Usage (MB-%)| 5.8 - 87.04 | 2 - 5.41 | 38.7 - 5.34 | 14 - 5.03 |
## [1539. Kth Missing Positive Number.][1539]
Go through array with a counter initialized to 1. If we encounter a value that
is equal to our counter (which denotes the strictly increasing sequence), we just
increment the counter, if we encounter a value larger than the counter, we adjust
`k` to account for the total gap between our counter and the value, and set our
counter to denote the next value after the value just encountered.
Complexity is `O(N)`, we perform constant operations for each element of the array.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 90.52 | 0 - 100.00 | 72 - 96.39 | 36 - 99.78 |
| Mem Usage (MB-%)| 6.1 - 97.71 | 2 - 96.67 | 38.1 - 99.78 | 14.3 - 69.55 |
## [1544. Make the string great.][1544]
We go through the string checking pairs. If a pair is bad (found by checking the
difference of their code point) we skip it, if not, we keep the first element
of the pair.
When we encounter a bad pair we need to take care to remove any bad pairs that
are created by the next character in the string and the last element we kept.
We therefore check if a bad pair exists between those two characters and, if not,
we simply break, if yes, we skip the character in the skip and remove the last
kept character. We continue this operation until the pair isn't bad.
Space and time complexity is O(N). Even while removing bad pairs, we skip characters
in the string.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 84 - 93.68 | 28 - 97.16 |
| Mem Usage (MB-%)| 5.8 - 44.44 | 2 - 83.33 | 41 - 38.42 | 14.1 - 77.79 |
## [1550. Three Consecutive odds.][1550]
Not many options here, go through array and count odds. Bail when you find them.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 97.43 | 40 - 89.59 |
| Mem Usage (MB-%)| 5.9 - 60.71 | 1.9 - 100.00 | 38.7 - 49.52 | 14.3 - 65.73 |
## [1556. Thousand Separator.][1556]
Make an array to hold result and a string out of the number (or, in `C`'s case,
simply a string). Then go through it backwards and insert separator
`'.'` as indicated by a counter that keeps track of numbers added.
`C` version is probably the more interesting.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 92.14 | 20 - 99.24 |
| Mem Usage (MB-%)| 5.5 - 87.50 | 2 - 88.89 | 38.5 - 71.43 | 14.1 - 86.04 |
## [1557. Minimum Number of Vertices to Reach All Nodes]
Return nodes with in-degree 0. Python uses sets to find them, the rest use an
array to count counts and then returns those for which `count == 0`.
Space and runtime complexity is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 276 - 100.00 | 24 - 100.00 | 172 - 96.46 | 1144 - 90.57 |
| Mem Usage (MB-%)| 38.1 - 33.33 | 9 - 37.50 | 62.3 - 92.04 | 53.3 - 30.91 |
## [1560. Most visited sector in a circular track.][1560]
The idea here is that we don't need to count everything. There may be many cycles
through the track which don't give us any information about which of the sections
was the most visited.
Instead, we need to look at the beginning and the end of the array to see where
we start and end.
If we start at 1 or end at `N`, all sections have been visited the same number
of times.
If start `!= 1` we start from an odd section. We keep track of these odd sections
(until `N` or until the array is exhausted) since they contribute `+1` to the
sections.
If end `!=N`, we end at an odd section, similarly to start, we keep track of
all sections from `1` until the end since they also contribute `+1` to the
number of sections.
Joining these two together, we can find our final result.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 97.12 | 36 - 96.11 |
| Mem Usage (MB-%)| 6.2 - 100.00 | 2.1 - 100.00 | 40.9 - 27.88 | 14.1 - 94.91 |
## [1572. Matrix Diagonal Sum][1572]
Go through the array and concurrently sum both diagonals. Remove
the middle element (to handle double counting) if the matrix has
an odd length. `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 12 - 100.00 | 0 - 100.00 | 68 - 97.94 | 96 - 99.21 |
| Mem Usage (MB-%)| 6.6 | 2.1 - 100.00 | 40.6 - 25.18 | 14.2 - 100.00 |
## [1582. Special Positions in a Binary Matrix.][1582]
Use a set to hold already occupied previously seen columns. Then iterate
through the rows looking for candidates.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 100.00 | 0 - 100.00 | 68 - 100.00 | 148 - 97.80 |
| Mem Usage (MB-%)| 7.5 - 5.26 | 2.1 - 83.33 | 40.4 - 91.67 | 14.4 - 92.64 |
## [1588. Sum of All Odd Length Subarrays][1588]
TODO: Explain the clusterfuck of the thing you call logic.
Note: Javascript perf is quite bad. Not sure what I'm doing
wrong or if another better trick exists. Surely, the mults array
doesn't *need* to be stored, we can compute it on the fly as
we go through indexes. Maybe I'll fix that in the future.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00% | 0 - 100.00% | 84 - 50.48% | 36ms - 94.20%|
| Mem Usage (MB-%)| 6 - 93.55% | 2 - 100.00% | 39 - 7.44% | 14 - 100.00%|
## [1598. Crawler Log Folder][1598]
Pretty straight-forward implementation of problem statement.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 90.48 | 0 - 100.00 | 76 - 91.88 | 40 - 90.93 |
| Mem Usage (MB-%)| 6.4 - 85.71 | 2.1 - 90.48 | 39 - 60.00 | 14.2 - 92.01 |
## [1603. Design Parking System][1603]
Mostly an OOP problem really.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 56 - 89.80 | 16 - 100.00 | 148 - 81.63 | 136 - 75.59 |
| Mem Usage (MB-%)| 20.8 - 100.00 | 2.4 - 45.00 | 46 - 26.12 | 14.6 - 74.11 |
## [1608. Special Array with X elements greater than or equal to X.][1608]
For each candidate special number (`1..array.length() + 1`) go through
the elements of the array and see if we have a matching number of elements
`>=` to it.
We can limit the iterations/comparisons made by breaking early when:
- matches are `>` than the special number.
- the remaining elements in the array aren't enough to reach the special number.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 97.50 | 36 - 74.54 |
| Mem Usage (MB-%)| 5.8 - 88.24 | 2 - 85.71 | 38.8 - 53.33 | 14 - 97.45 |
## [1614. Maximum Nesting Depth of the Parentheses][1614]
Traverse string while keeping count of depth. Runtime `O(n)` and
space `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.16 | 24 - 95.18 |
| Mem Usage (MB-%)| 5.7 - 85.33 | 2.2 - 100.00 | 39.2 - 76.73 | 14.1 - 100.00 |
## [1619. Mean of Array after removing some elements.][1619]
Sort, slice and return mean.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 76 - 97.64 | 48 - 97.10 |
| Mem Usage (MB-%)| 6.1 - 91.67 | 2.1 - 85.19 | 40.3 - 29.06 | 14.1 - 95.31 |
## [1624. Largest substring between two equal characters.][1624]
Create a map holding min and max value for a given character in `s`. Then, go
through the map values and find the max distance.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 91.62 | 24 - 95.74 |
| Mem Usage (MB-%)| 6 - 10.36 | 2 - 100.00 | 39.5 - 22.16 | 14.3 - 10.05 |
## [1629. Slowest Key.][1629]
Find and keep all indices that have max `releaseTimes[i] - releaseTimes[i-1]` in a list.
Then go through that array and find the max character to return.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 92.16 | 0 - 100.00 | 76 - 92.58 | 48 - 96.71 |
| Mem Usage (MB-%)| 6.2 - 96.08 | 2.1 - 72.22 | 39.9 - 54.42 | 14.6 - 22.01 |
## [1636. Sort Array by Increasing Frequency.][1636]
Need to build counts and then perform a two-way sort. First we sort by
value (stability isn't a concern) and then by frequency.
Stability when sorting by frequency is a concern because we want to reverse
the order of elements with the same frequency. To do this we need to
add a tie-braker (the position after the first sort) to use when sorting
again by position.
Since the second sort results in the array being reversed, we need to add
elements from the end towards the front.
**TODO: JS sort happens to be implemented as stable (this isn't guaranteed
though, need to use same tie-braking mechanism as C there.**
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 84 - 98.23 | 40 - 98.75 |
| Mem Usage (MB-%)| 7.3 - 28.00 | 2 - 71.43 | 41.8 - 23.89 | 13.9 - 97.38 |
## [1637. Widest vertical area between two points containing no points.][1637]
We don't need the `y` axis to find greatest vertical distance, we need the `x` axis
in sorted order. We can then find the max difference between two adjacent points.
Note: Though it doesn't help much (at least not in any besides Python), we can also
notice that we need distinct `x` values. We can use a set to filter duplicate x values
while building the array holding them.
This approach is only used in Python, in others, it didn't result in any benefit.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 200 - 92.00 | 24 - 90.00 | 172 - 98.66 | 800 - 95.57 |
| Mem Usage (MB-%)| 28.6 - 64.00 | 9.6 - 96.00 | 62.6 - 32.89 | 55.3 - 19.34 |
## [1640. Check Array Formation Through Concatenation][1640]
Build a set-like supporting structure for fast O(1) look-ups.
Go through each element in second array:
- If the sub-array has a single element, check if its contained
in the set-like structure and bail if not.
- If the sub-array has more than one elements. Find index of
subarray[0] in original array and try and match the rest of
the values in the sub-array.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 92.31 | 0 - 100.00 | 80 - 67.57 | 36 - 93.55 |
| Mem Usage (MB-%)| 6.7 - 100.00 | 2.2 - 100.00 | 38.8 - 5.11 | 14.1 - 100.00 |
## [1641. Count sorted vowel strings.][1641]
The verbose initial solution can be seen in the Python file. There, a dictionary is created to
hold the values. After closer examination I realized a mapping is really not needed when we
have 3-4 variables we need to keep track of.
For each of the characters, for step `n+1` its value is determined by adding the sum of the next
character in the sequence to what that character already holds. In essence:
```python
u = 1
o += u
i += o
e += i
a += a
```
Using the previous relation in a while loop gets us our result (the value of `a`). `O(N)` runtime and
`O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.77 | 20 - 99.64 |
| Mem Usage (MB-%)| 5.5 - 65.12 | 1.9 - 92.31 | 38.1 - 98.62 | 14.4 - 30.55 |
## [1646. Get maximum in generated array.][1646]
Initially thought there was a trick in order to immediately find the value in the
array for a given `n`. According to the results, I doubt that now.
Build the array and while building hold onto the max seen.
Space/Time complexity is `O(n)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 93.76 | 24 - 96.72 |
| Mem Usage (MB-%)| 5.9 - 11.17 | 2 - 95.00 | 38.9 - 10.99 | 14 - 99.22 |
## [1652. Defuse the bomb.][1652]
Use modular arithmetic (dividing by array lenght) to add correct elements.
In Python, use `deque` to rotate around instead.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 97.14 | 0 - 100.00 | 72 - 98.31 | 36 - 84.77 |
| Mem Usage (MB-%)| 6.1 - 85.71 | 2 - 100.00 | 38.6 - 84.46 | 14.1 - 89.93 |
## [1656. Design An Ordered Stream][1656]
There's probably a different way for this. I've created a
mapping to hold id-value pairs and after every insert I
simply create the resulting array by increasing ptr and
checking if a key with that value exists in the mapping.
Pretty sure there's got to be a simpler way.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 84 - 62.50 | 20 - 93.33 | 184 - 89.38 | 200 - 98.57 |
| Mem Usage (MB-%)| 33.5 - 12.50 | 2.9 - 40.00 | 49.5 - 100.00 | 14.7 - 75.25 |
## [1662. Check if two string arrays are equivalent.][1662]
Go through characters and check, alternatively, build resulting string and
compare the string.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 91.70 | 24 - 97.07 |
| Mem Usage (MB-%)| 5.9 - 91.49 | 2 - 80.00 | 38 - 99.79 | 14 - 96.79 |
## [1672. Richest customer wealth.][1672]
Max of the sum of the entries.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0- 100.00 | 72 - 92.65 | 48 - 95.26 |
| Mem Usage (MB-%)| 6.1 - 94.74 | 2 - 66.67 | 38.4 - 83.19 | 14.1 - 86.41 |
## [1678. Goal Parser Interpretation.][1678]
Go through the string via index which we increase based on the chunk we've seen.
For C, we can alter the input string instead of creating a new one.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 100.00 | 24 - 98.37 |
| Mem Usage (MB-%)| 5.8 - 70.00 | 2 - 100.00 | 38.5 - 26.43 | 14.1 - 72.89 |
## [1684. Count the number of consistent strings.][1684]
Create a set of the allowed characters and go through array checking if each
character of each word is inside the set.
Feel (mainly due to `C` timings) that there must be something I'm not taking
advantage of here.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 48 - 69.33 | 12 - 97.06 | 108 - 93.83 | 224 - 85.91 |
| Mem Usage (MB-%)| 12.3 - 36.00 | 2.7 - 17.65 | 47.8 - 80.74 | 16 - 95.99 |
## [1688. Count of matches in tournament.][1688]
Add the divmod of `n` by `2` continuously until `n` reaches `1` (last match).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 85.89 | 16 - 99.96 |
| Mem Usage (MB-%)| 5.5 - 78.38 | 2 - 91.18 | 38.5 - 76.57 | 14.2 - 45.53 |
## [1689. Partition into minimum number of deci-binary numbers.][1689]
Idea is simple after you view digits as points in a vector (at least,
that's how I visualized it).
We can continuously reduce by one each position until it reaches zero,
no matter what the number is, at each step every single one of its positions
will be reduced by one. As an example with `n = "93240032"`:
We can now create a deci-binary number that has a 1 for each position in n > 0
and a zero if the position in n is 0:
```
"93240032"
- "11110011" <-- deci-binary
= "82130021"
```
This process can continue untill all the digits of the number are equal to zero.
This leads to the minimal set of numbers since it is the minimal set of ones to
reach the max digit (we can only increase by one, there's no faster way to reach
a given digit.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 88 - 91.16 | 44 - 96.88 |
| Mem Usage (MB-%)| 7.8 - 83.33 | 2.1 - 59.46 | 42.8 - 95.98 | 14.6 - 97.73 |
## [1694. Reformat phone numbers.][1694]
For `Python` and `Rust`: First remove all `'-'` and `' '` characters from the
string and then iterate through chunks which we save to a result array. Then
we return the array joined on `'-'`.
For `C` and `JavaScript`: Go through character by character carefully collecting
numbers and placing the `'-'` character where needed, see comments in source
files.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 91.67 | 28 - 93.41 |
| Mem Usage (MB-%)| 5.7 - 83.33 | 2 - 100.00 | 40.1 - 8.33 | 14.2 - 94.48 |
## [1700. Number of students unable to eat lunch.][1700]
Use a (de)Queue to efficiently pop from the left and rotate. With it, the
implementation is straight-forward.
For `C`: When I (finally) build a little Queue object.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| N/A | 0 - 100.00 | 80 - 100.00 | 28 - 100.00 |
| Mem Usage (MB-%)| N/A | 2.3 - 100.00 | 40 - 100.00 | 14.2 - 100.00 |
## [1704. Determine if string halves are alike.][1704]
Build a count of vowels until the midpoint of the string. Then go through the string
from the midpoint until the end and reduce the count. If at any point it becomes
negative, we're done, else, return result of `count == 0`.
A set is mostly used to hold the vowels, except for `C` in which we keep them in an
array and iterate through it.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 99.53 | 24 - 99.54 |
| Mem Usage (MB-%)| 6 - 41.38 | 2 - 100.00 | 38.9 - 81.04 | 14.1 - 89.35 |
## [1710. Maximum Units on a Truck.][1710]
Greedy solution, we continuously select boxes with maximum units until no more boxes
can fit inside the truck.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 28 - 100.00 | 4 - 100.00 | 96 - 100.00 | 144 - 100.00 |
| Mem Usage (MB-%)| 7.7 - 100.00 | 2.2 - 100.00 | 42.1 - 100.00 | 14.9 - 100.00 |
## [1716. Calculate Money in Leetcode Bank.][1716]
Can go through `n` in weeks and then add left-over days.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 96.12 | 24 - 98.92 |
| Mem Usage (MB-%)| 5.5 - 84.44 | 2 - 100.00 | 38.3 - 61.79 | 14.2 - 71.27 |
## [1720. Decode xored array.][1720]
Based on xor-swap algorithm. Can find the original value used to get the xor'ed
values simply by xor-ing the new values.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 100 - 100.00 | 8 - 100.00 | 112 - 94.12 | 216 - 98.72 |
| Mem Usage (MB-%)| 20.7 - 62.50 | 2.2 - 100.00 | 45 - 65.55 | 15.7 - 98.32 |
## [1725. Number of Rectangles that can form the largest square.][1725]
Note that the largest square has a side that is equal to the maximum between the minimum
of the pairs inside the array.
To find the count, we hold a max value and count value initialized to zero. For each
pair inside the array, if the minimum of the pair is equal to max, we just increase
the count (signifying that we've found another square with `side == max`).
If the minimum of the pair is larger than the current max we've currently counted,
we need to reset our counter to 1 (to count the current square with side `> max`)
and set `max` to the new maximum.
At the end, we have the count of the squares with sides equal to the maximum value
possible.
Runtime is `O(N)` since we perform constant operations for each iteration through the
array. Space complexity is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 36 - 100.00 | 4 - 100.00 | 92 - 100.00 | 172 - 100.00 |
| Mem Usage (MB-%)| 8 - 100.00 | 2.2 - 100.00 | 41.5 - 100.00 | 14.8 - 100.00 |
## [1732. Find the highest altitude.][1732]
Initialize max and height to zero. Go through gains and add to height, if height is
larger than max, set max to height.
Return max.
Runtime O(N), space O(1).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 64 - 100.00 | 28 - 100.00 |
| Mem Usage (MB-%)| 6 - 100.00 | 2.1 - 100.00 | 38.6 - 100.00 | 14.1 - 100.00 |
## [1742. Maximum Number of Balls in a box][1742]
Go through range counting indices. Best trick is to not compute the indices for
all numbers. Rather take advantage of fact that for a given number, we can compute
the base index of its multiple of ten and then add to that the next nine values
from [0-9].
For example, given ranges `[39, 73]`, we can skip computing indices for all numbers
and instead compute them for `[30, 40, 50, 60, 70]`, i.e multiples of ten.
The remaining numbers can be added by adding the
range from `[0, 9]` to that base number, special cases apply for the initial multiple
of ten and the final one (because we have remainders). Coarsely, what we compute is:
```python
# 3 is quotent 30 // 10
sum_digits(3) -> 3
# 9 is the remainder of 39 % 10.
add_to_counter_with_base(3, 9, 10)
# 4 is quotent of 40 // 10
sum_digits(4) -> 4
# add full range from 0-9.
add_to_counter_with_base(4, 0, 10)
# 5 is quotent of 50 // 10
# add full range from 0-9
add_to_counter_with_base(5, 0, 10)
# similarly for 60
# seven is quotent of 70 // 10.
sum_digits(7) -> 7
# 3 is remainder of 73 % 10
add_to_counter_with_base(7, 0, 3)
```
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 92 - 99.58 | 156 - 99.80 |
| Mem Usage (MB-%)| 5.6 - 58.82 | 2 - 59.09 | 38.6 - 90.72 | 14.4 - 33.47 |
## [1748. Sum of unique elements][1748]
Rust, Javascript, Python all use sets in order to keep track of seen numbers and
unique numbers. All could use `C`s alternative of using a bit-array in order to
keep track of these and reduce memory requirements. All in all, all run in `O(N)`
and while `C`s memory complexity should also be `O(N)` it definitely is a much
smaller one.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 84 - 100.00 | 32 - 100.00 |
| Mem Usage (MB-%)| 5.8 - 100 | 2.1 - 100.00 | 39.1 - 100.00 | 14.3 - 100.00 |
## [1752. Check if array is sorted and rotated.][1752]
Keep track of starting value in nums.
Iterate through `nums` checking if the invariant (`nums[i-1] < nums[i]`) holds. If we
reach the end, we have a sorted array with no rotations.
If we don't reach the end (i.e `nums[i-1] > nums[i]`), we've reached a possible point
where the array was rotated. In order for the array pre-rotation to be valid two
things need to hold:
- same invariant (i.e `nums[i-1] < nums[i]`)
- all values must be `<= start`.
If the first condition doesn't hold, the pre-rotated array wasn't sorted in non-decreasing
order.
If the second condition doesn't hold, the pre-rotated array wasn't sorted since `start`
is smaller than an element that is found to its left.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 100.00 | 24 - 100.00 |
| Mem Usage (MB-%)| 6.1 - 100.00 | 2.1 - 100.00 | 38.4 - 100.00 | 14 - 100.00 |
## [1768. Merge Strings Alternatively.][1768]
Zip strings and build until length of smallest. Then push what's left from the largest
to the result.
O(N) where N is size of largest string.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 60 - 99.88 | 24 - 95.91 |
| Mem Usage (MB-%)| 5.7 - 100.00 | 2.1 - 60.42 | 39.2 - 20.05 | 14.1 - 81.02 |
## [1773. Count items matching a rule.][1773]
Find index which `ruleKey` corresponds to and then count items matching the `ruleValue`. This results in `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 60 - 100.00 | 12 - 100.00 | 84 - 100.00 | 240 - 100.00 |
| Mem Usage (MB-%)| 13.6 - 100.00 | 4.4 - 100.00 | 42.4 - 100.00 | 20.7 - 50.00 |
## [1779. Find nearest point that has the same x or y coordinate.][1779]
`O(N)`, the main insight (i guess) is the fact that you don't really need to calculate one part of the manhattan distance
since you know it will be zero (`xi == x <=> xi - x == 0`, similarly for `yi, y`).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 140 - 98.46 | 16 - 97.73 | 104 - 94.58 | 688 - 97.23 |
| Mem Usage (MB-%)| 16.1 - 61.54 | 2.8 - 100.00 | 46.3 - 88.14 | 19.5 - 13.48 |
## [1790. Check if one string swap can make strings equal.][1790]
Check each of the characters in both string to verify that they differ in less than 2 positions (`O(N)`); if
there's more than two mismatches in position, we'd need more than two swaps.
After doing that we need to check that the possible switch involves the same characters. This could probably be done
by keeping track of the indices of the mismatch but I've done it by using a set equality between strings (`O(N)`).
Overall, runtime complexity is `O(N)` while space complexity is `O(1)` (because we only deal with 26 different characters
as input, the size of the sets is always bound by that.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 93.86 | 24 - 94.90 |
| Mem Usage (MB-%)| 5.4 - 93.26 | 2.1 - 55.17 | 40.2 - 20.55 | 14.3 - 20.60 |
## [1791. Find center of star graph.][1791]
Find the common node in `edges[0]` and `edges[1]`. `O(1)` space/time.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 168 - 92.23 | 16 - 95.45 | 108 - 96.41 | 760 - 97.72 |
| Mem Usage (MB-%)| 25.6 - 98.06 | 8.8 - 100.00 | 51.8 - 96.65 | 50 - 96.41 |
## [1796. Second largest digit in a string.][1796]
Go through digits and find the second maximum value. Pretty straight-forward, `O(N)` runtime with
`O(1)` space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 98.50 | 28 - 97.85 |
| Mem Usage (MB-%)| 5.7 - 82.69 | 2 - 100.00 | 39.1 - 97.50 | 14 - 93.74 |
## [1797. Design Authentication manager.][1797]
Use a map to handle the tokens and with a slight amount of logic keep them updated. Feels
like I'm missing something obvious here (e.g a better way to store data in order to not
hold on to stale-expired tokens).
All ops except for `count_unexpired`, which is `O(N)`, are `O(1)`
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 20 - 100.00 | 160 - 93.33 | 246 - 71.63 |
| Mem Usage (MB-%)| - | 3.3 - 100.00 | 48.2 - 80.00 | 15.5 - 83.72 |
## [1800. Maximum Ascending Subarray sum.][1800]
Keep track of current sum and the max sum we've found so far. When we detect the end of a sequence, we just update the
max sum if the current sum is larger. Care is needed to handle last element in the list of numbers. Overall, `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 90.78 | 24 - 99.12 |
| Mem Usage (MB-%)| 5.8 - 75.00 | 2 - 96.15 | 38.5 - 55.32 | 14.2 - 75.50 |
## [1805. Number of different integers in a string.][1805]
Split/trim strings and use a set to count unique elements. `C` version shouldn't have
been difficult just a bit of a pain. All in all, leads to `O(N)` execution and spacetime
complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 72 - 96.63 | 20 - 99.50 |
| Mem Usage (MB-%)| - | 0 - 100.00 | 38.5 - 95.04 | 14.4 - 27.93 |
## [1812. Truncate sentence.][1812]
Observe that if we map cols from `[a-h]` to `[1-8]` we have a nice property: cells for which
the parity is the same are black while those for which parity is different are white. Using this,
solution is `O(1)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 95.14 | 24 - 96.62 |
| Mem Usage (MB-%)| 5.4 - 75.00 | 1.9 - 96.55 | 38.4 - 78.42 | 14.1 - 90.58 |
## [1816. Truncate sentence.][1816]
Simple `O(N)`, enumerate through chars counting spaces, when k reaches zero, return the index for that
iteration. Return string splitted/truncated on that index.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 94.14 | 20 - 99.16 |
| Mem Usage (MB-%)| 5.9 - 73.68 | 2 - 67.50 | 39.1 - 6.25 | 14 - 99.43 |
## [1822. Sing of the product of the array.][1822]
No need to do the actual multiplications. Simply toggle a variable the amount of times a negative number is seen. If
zero is encountered, zero is unconditionally returned. `O(N)` since we examine all values.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 92.86 | 56 - 93.29 |
| Mem Usage (MB-%)| 6.3 - 35.90 | 2.1 - 53.33 | 39.8 - 47.25 | 14.4 - 70.94 |
## [1827. Minimum operations to make the array increasing.][1827]
Step through array in pairs and increment by one if equal and by their difference + 1 if
earlier pair is largest. `O(n)` time, `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 98.73 | 0 - 100.00 | 72 - 97.45 | 116 - 96.62 |
| Mem Usage (MB-%)| 6.5 - 59.49 | 2 - 92.31 | 39.9 - 94.91 | 15 - 83.39 |
## [1832. Check if the sentence is pangram.][1832]
Use a set to filter characters and check that set size is equal to 26.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 97.22 | 16 - 99.92 |
| Mem Usage (MB-%)| 5.6 - 97.71 | 2 - 97.14 | 40.4 - 13.63 | 13.9 - 99.13 |
## [1837. Sum of digits in base k.][1837]
Find the digits of the representation in base-k by taking `n mod k`. Sum the digits and
add whatever remains in the end of `n` (where `n` will be `< k`).
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 100.00 | 24 - 100.00 |
| Mem Usage (MB-%)| 5.4 - 100.00 | 2 - 100.00 | 38.6 - 100.00 | 14 - 100.00 |
## [1844. Replace all digits with characters.][1844]
Straight-forward traversal through the characters of the input string. `O(n)` time
complexity and `O(N)/O(1)` space complexity for languages that don't and do allow
mutation of strings respectively.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 99.54 | 24 - 97.22 |
| Mem Usage (MB-%)| 5.6 - 88.89 | 2 - 95.24 | 38.5 - 60.19 | 14.2 - 43.57 |
## [1848. Minimum distance to the target element.][1848]
Iterate through the list and find the value whose index yields the minimum difference.
`O(N)` time and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 91.67 | 0 - 100.00 | 72 - 94.62 | 48 - 96.71 |
| Mem Usage (MB-%)| 6 - 90.28 | 2 - 91.89 | 38.8 - 50.14 | 14.2 - 14.2 - 97.36 |
## [1854. Maximum population year.][1854]
Iterate through all the logs and build a counter that holds number of people alive for each
year. After that, we only need to traverse this array to find max value for min year.
Since logs are bounded in size (maximum 100 elements), the inner loop while iterating through
the logs is basically a constant term in the complexity (similarly, the counter used to count
occurences). It does not depend on the input size `N`. As such, this is `O(N)` runtime and `O(1)`
space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 97.89 | 32 - 98.79 |
| Mem Usage (MB-%)| 6.3 - 27.50 | | 39 - 71.90 | 14.2 - 65.92 |
## [1859. Sorting the sentence.][1859]
Get number of strings by calling len after splitting. Pre-allocate vector to hold
strings and put strings as encountered in correct position. O(n) both space and time.
`C` too much hassle for me to get a join implemented now.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 72 - 97.39 | 28 - 92.23 |
| Mem Usage (MB-%)| - | 2.1 - 58.33 | 38.4 - 52.29 | 13.9 - 98.94 |
## [1863. Sum of all subset xor totals.][1863]
View subsequences as bitstrings where bit set
indicates we should pick a specific index.
We essentially generate the following sequence:
0, 01, 10, 11, 001, 010, 100, 011, 101, 110, 111
and use that to index into the array (via `k & (1 << i)`)
which checks if the the i'th bit is set.
If I'm not mistaken, this trick is described in Knuth's most recent book (4A?)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 4 - 100.00 | 80 - 100.00 | 152 - 100.00 |
| Mem Usage (MB-%)| 5.4 - 100.00 | 2 - 100.00 | 38.4 - 100.00 | 14.1 - 100.00 |
## [1876. Substrings of size three with distinct characters.][1876]
Go through the string in pairs of 3s and check them. `O(N)` runtime and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 68 - 99.35 | 24 - 99.11 |
| Mem Usage (MB-%)| 5.5 - 95.00 | 1.9 - 100.00 | 39.3 - 79.08 | 14 - 89.81 |
## [1880. Check if word equals summation of two words.][1880]
Transform strings into ints as described and check the equality. Nothing fancy here.
Has `O(N)` time and space complexity.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 80 - 100.00 | 20 - 100.00 |
| Mem Usage (MB-%)| 5.8 - 100.00 | 2 - 100.00 | 38.4 - 100.00 | 14.1 - 100.00 |
## [1893. Check if all the integers in a range are covered.][1893]
Sort the input ranges and then step through them. Whenever we encounter a range in which
`left` is contained, advance left to the end of the range `+ 1`. Similarly for `right` with
the difference that we reduce until the start of the range `-1`. This is done until `left > right`
or we've run out of ranges.
In the end, if `left > right`, we've covered our range. `O(nlogn)` runtime due to sorting, `O(1)` extra
space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 64 - 100.00 | 28 - 99.93 |
| Mem Usage (MB-%)| 6.2 - 100.00 | 2 - 100.00 | 40.1 - 18.02 | 14.3 - 60.94 |
## [1897. Redistribute characters to make all strings equal.][1897]
Build a count of the characters and check, for each character count, that `count % words_length`
is equal to zero. If it is, all characters can be evenly distributed with no odd chars left out.
Space is constant (same array regardless of input) while time complexity is `O(N*s)` where `N`
is the number of elements in the array and `s` is the string size.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 80 - 99.25 | 48 - 99.40 |
| Mem Usage (MB-%)| 6.9 - 31.25 | 2 - 100.00 | 40.8 - 92.48 | 14.3 - 89.13|
## [1902. Largest odd number in string.][1902]
Make a little set of the odd numbers (or a ten element array to use as a set) and go through
the string backwards. When we first encounter an odd number, we return.
Time complexity is `O(N)` while space complexity is constant (except for Rust where I `.cloned()` the
string to make my life easier.)
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 100.00 | 0 - 100.00 | 84 - 100.00 | 32 - 100.00 |
| Mem Usage (MB-%)| 8.6 - 100.00 | 2.4 - 100.00 | 42.6 - 100.00 | 15.3 - 20.00 |
## [1908. Remove one element to make the array strictly increasing.][1908]
Traverse array looking for a pair `nums[i], nums[i+1]` for which `nums[i] >= nums[i+1]`. Now, we need
to decide which of the two values should be removed and continue checking. If we've already removed a
value at some point in the past, we can bail immediately.
To choose which of `a (nums[i])` or `b (nums[i+1])` we should remove, we look at the value left of `a` and
the value right of `b`. Removals are basically tracked with a boolean flag and what we remove is controlled
by our counter.
If the value left of `a` is larger than `b`, we remove `b` if the value on the right
of `b` is larger than `a` (meaning we'd have a sequence like `4, 5, 1, 7`). If the value on the right of `b`
isn't larger, we bail, nothing can be done in this case (i.e `(5, 10, 1, 2)`).
If the value left of `a` isn't larger than `b` then `a` is removed.
All in all, this is `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 81.82 | 0 - 100.00 | 68 - 98.43 | 44 - 96.90 |
| Mem Usage (MB-%)| 5.9 - 100.00 | 2 - 60.00 | 39.4 - 66.93 | 14.5 - 50.11 |
## [1912. Maximum product difference between two pairs.][1912]
Find 2 largest (`a, b`) and 2 smallest (`c, d`) and return `a * b - c * d`. `O(N)` where `N` is
size of the array.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 20 - 98.33 | 0 - 100.00 | 80 - 94.71 | 160 - 94.50 |
| Mem Usage (MB-%)| 7.2 - 65.00 | 2.1 - 50.00 | 41.5 - 79.74 | 15.4 - 83.12 |
## [1920. Build array from permutation.][1920]
I'm building a new array here, so `O(N)` both space and time complexity.
TODO: Tried for over an hour to work out how an `O(1)` approach would work. Definitely
uses some index tricks but I can't figure it out. Need to check it again in the futuro.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 40 - 100.00 | 4 - 100.00 | 100 - 100.00 | 112 - 100.00 |
| Mem Usage (MB-%)| 10 - 100.00 | 2 - 100.00 | 43 - 100.00 | 14.4 - 33.33 |
## [1925. Count square sum triples.][1925]
Precompute `n` array of powers from `1..n`. Loop through (`i` counter) for
all powers from `3..n-1`, then through the powers from `i..n` and try and
see if you get a value that maches the expected result.
That can be done in one of two ways:
1. Compute `sqrt(pows[i] + pows[j])` and check if it is a perfect squate.
2. Use a set and check if `pows[i] + pows[j]` is in that set.
The first approach appears to be the fastest.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 3 - 88.89 | 2 - 88.00 | 68 - 98.38 | 136 - 94.74 |
| Mem Usage (MB-%)| 5.8 - 22.22 | 1.9 - 92.00 | 40.9 - 19.46 | 14.1 - 77.35 |
## [1926. Nearest exit from entrance in maze.][1926]
Use breadth first search to find the shortest exit.
C: Fixing deque and a set/2d array to hold visited positions is a hassle.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 48 - 100.00 | 120 - 89.02 | 780 - 95.42 |
| Mem Usage (MB-%)| - | 3.3 - 33.33 | 49.9 - 35.37 | 15.6 - 41.52 |
## [1929. Concatenation of Array.][1929]
Yeah, not really challenging. Just extend the nums array. `O(N)`.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 32 - 100.00 | 4 - 100.00 | 100 - 100.00 | 76 - 100.00 |
| Mem Usage (MB-%)| 9.9 - 100.00 | 2.1 - 100.00 | 41.6 - 100.00 | 14.4 - 100.00 |
## [1935. Maximum number of words you can type.][1935]
Simple counting problem. `O(N)` time and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 100.00 | 28 - 100.00 |
| Mem Usage (MB-%)| 5.6 - 100.00 | 2 - 100.00 | 40.4 - 100.00 | 14.2 - 100.00 |
## [1941. Check if all characters have equal number of occurences.][1941]
Again a simple counting problem, `O(N)` time and `O(1)` space.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 76 - 100.00 | 24 - 100.00 |
| Mem Usage (MB-%)| 6 - 100.00 | 2.1 - 100.00 | 40.2 - 100.00 | 14.2 - 66.66 |
## [1945. Sum of digits of string after convert.][1945]
Lots of conversions but not something really tricky. Need to take care to always work
with individual digits. `O(N)` time and `O(N)` space.
C: Not up for getting into the hairy details of constant transformation between chars
and digits which we need to store in arrays.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 0 - 100.00 | 80 - 100.00 | 32 - 100.00 |
| Mem Usage (MB-%)| | 2.1 - 100.00 | 41.6 - 100.00 | 14.2 - 100.00 |
## [1961. Check if string is a prefix of an array.][1961]
Can either build the prefixes until we match with the given string or gradually
remove prefixes from the string until it is empty.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 36 - 93.75 |
| Mem Usage (MB-%)| 5.6 - 100.00 | 2.1 - 66.67 | 2.4 - 94.74 | 13.9 - 67.19 |
## [1967. Number of strings that appear as substrings in word.][1967]
Use `in`, `string.Contains`, `String.contains` and `strstr` to check if the substring
is in the word. Should be `O(N*M)` where `N` is the number of words and `M` is the length of the
string.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 31 - 98.97 |
| Mem Usage (MB-%)| 2.2 - 37.50 | 2 - 83.33 | 2.5 - 97.67 | 13.8 - 79.47 |
## [1974. Minimum time to type word using special typewriter.][1974]
End result is going to be `len(s) + rotations`. `len(s)` signifies the seconds needed to print
out each individual characters. `rotations` is calculated as follows:
For each character `c` is `s` and the next character `c2` in s, we need to find
`minimum(c->c2, c2->c)` where the arrows signify distance between `c` and `c2` clockwise and
counterclockwise.
This is generally easy since we're in a cycle. `c->c2` is simply `|ord(c) - ord(c2)|` while `c2->c` will
be `26 - |ord(c) - ord(c2)|` (since we have 26 characters).
In addition to the previous, we need to handle the special case that the word doesn't start with an `'a'`.
All in all, the operation is `O(N)` with `O(1)` space requirements.
| Stats/Lang | C | Rust | JS | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 72 - 89.33 | 24 - 97.62 |
| Mem Usage (MB-%)| 5.5 - 86.67 | 2.1 - 69.23 | 38.3 - 98.00 | 14.1 - 73.63 |
## [1979. Find greatest common divisor of array.][1979]
Find min/max by iteration and then just calculate the GCD using Euclid's algorithm.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 4 - 96.88 | 0 - 100.00 | 3 - 97.56 | 60 - 93.13 |
| Mem Usage (MB-%)| 6.2 - 22.92 | 2.2 - 7.14 | 3 - 100.00 | 13.9 - 98.13 |
## [2000. Reverse prefix of word.][2000]
Pretty straightforward. For C, which allows in-place manipulation, we have `O(N)` time and `O(1)` space, for
the rest, we have `O(N)` time and `O(N)` space.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 34 - 88.86 |
| Mem Usage (MB-%)| 5.7 - 57.58 | 2.2 - 33.33 | 2 - 92.86 | 13.8 - 63.95 |
## [2006. Count Number of Pairs with absolute difference k.][2006]
Create a counter for all the numbers in `nums`. Then loop through all the numbers in `nums` and check if
`num - k` or `num + k` is in it. Add the count for each to the total count and decrement the count for the
`num` we just checked. This is `O(N)` time and `O(N)` space.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 92.22 | 0 - 100.00 | 5 - 91.03 | 66 - 97.16 |
| Mem Usage (MB-%)| 5.8 - 78.89 | 2.1 - 88.24 | 4.1 - 16.67 | 13.8 - 98.06 |
## [2011. Final Value of Variable After Performing Operations][2011]
Python, Rust use functional style, C and Go the unwrapped loop. All are O(N) since a constant
ammount of ops is done for each operation.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 5 - 79.28 | 0 - 100.00 | 0 - 100.00 | 56 - 91.04 |
| Mem Usage (MB-%)| 6.8 - 8.84 | 2.1 - 93.42 | 3.5 - 15.61 | 13.8 - 56.21 |
## [2037. Minimum number of moves to seat everyone.][2037]
Sort both input arrays and then iterate over them pairwise and sum. `O(NlogN)`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 7 - 86.54 | 0 - 100.00 | 5 - 93.44 | 71 - 87.34 |
| Mem Usage (MB-%)| 5.9 - 100.00 | 2.1 - 60.00 | 3.3 - 85.25 | 13.8 - 97.72 |
## [2042. Check if numbers are ascending in a sentence.][2042]
Go through parsing the integers and compare, most of the hassle is in the locating/parsing of
the numbers.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 34 - 90.10 |
| Mem Usage (MB-%)| 5.8 - 23.53 | 2 - 100.00 | 1.9 - 96.97 | 14 - 11.64 |
## [2057. Smallest index with equal values.][2057]
Pretty straight-forward, I do feel like I'm missing some trick, though.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 10 - 82.50 | 0 - 100.00 | 7 - 97.50 | 87 - 95.09 |
| Mem Usage (MB-%)| 6 - 70.00 | 2.1 - 66.67 | 4.2 - 67.50 | 13.9 - 8.83 |
## [2089. Find target indices after sorting array.][2089]
Sort array and iterate, straight forward. `O(NlogN)`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 3 - 93.40 | 0 - 100.00 | 0 - 100.00 | 48 - 95.31 |
| Mem Usage (MB-%)| 5.8 - 95.28 | 2.1 - 61.90 | 2.8 - 81.08 | 13.8 - 97.69 |
## [2103. Rings and Rods.][2103]
Create a vector of rods and for each, each value will denote a bitmap of the rings that
are placed on it. Then, for each rod, we check if the value of the bitmap is equal to of
all rings or'ed together, if so, we increment the count.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 33 - 92.33 |
| Mem Usage (MB-%)| 5.7 - 54.17 | 2 - 100.00 | 1.9 - 95.45 | 13.9 - 64.60 |
## [2108. Find First Palindromic String in the Array.][2108]
Go through the array and bail early when you find a palindrome. Relatively straight-forward.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 30 - 98.04 | 3 - 100.00 | 23 - 75.82 | 70 - 97.30 |
| Mem Usage (MB-%)| 9.8 - 35.29 | 2.1 - 94.12 | 7.6 - 31.87 | 13.9 - 68.33 |
## [2114. Maximum Number of Words Found in Sentence.][2114]
Iterate, split, count.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 3 - 99.09 | 0 - 100.00 | 3 - 86.74 | 49 - 83.94 |
| Mem Usage (MB-%)| 6.7 - 15.00 | 2.1 - 83.54 | 4.1 - 42.80 | 13.8 - 99.06 |
## [2119. A number after a double reversal.][2119]
Check if last digit is a zero (which would get ignored), edge case is `num == 0`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 33 - 91.09 |
| Mem Usage (MB-%)| 5.5 - 51.89 | 2.2 - 20.00 | 1.9 - 12.24 | 13.7 - 95.03 |
## [2124. Check if all a's appear before all b's.][2124]
Go through until meeting first `'b'` and then using a flag check if there are any `'a'`s after that.
Alternatively, for cases where we can (`C` and `Rust`) we can advance the string/iterator and then
re-advance after finding the `'b'` and breaking if we find an `'a'`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 37 - 85.22 |
| Mem Usage (MB-%)| 5.6 - 41.03 | 2 - 91.67 | 2 - 75.76 | 13.9 - 52.26 |
## [2154. Keep multiplying found values by two.][2154]
Build a set and continuously multiply while original is found in it, `O(N)` worse case.
C: Too bored to build the set.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| | 0 - 100.00 | 0 - 100.00 | 55 - 99.67 |
| Mem Usage (MB-%)| | 2.1 - 30.00 | 4.5 - 27.08 | 14 - 57.70 |
## [2160. Minimum sum of four digit number after splitting digits.][2160]
Split digits, sort and add 1st with 3rd and 2nd with 4th (minimum sum will always
be found by starting both splits with smallest possible digit)
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 33 - 90.09 |
| Mem Usage (MB-%)| 5.6 - 48.57 | 2.1 - 40.00 | 1.9 - 11.72 | 13.9 - 56.28 |
## [2169. Count operations to obtain zero.][2169]
Continuously reduce numbers while they are both larger than `1`. If one drops to zero, return the
number of ops we keep track of, else return the sum of the ops with whatever remains of the number
not equal to one.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 60 - 88.36 |
| Mem Usage (MB-%)| 5.6 - 16.20 | 2 - 76.92 | 1.9 - 53.33 | 13.9 - 63.79 |
## [2176. Count equal and divide pairs.][2176]
Python only for now, its late: created zipped array of values with their indices and sort. Go through
the array and check if the value is equal to the previous value and if the index difference is divisible
by `k`. If so, increment the count. Worse case, this is `O(N^2)`.
TODO: Do the rest.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | - | - | 68 - 98.17 |
| Mem Usage (MB-%)| - | - | - | 13.9 - 71.66 |
## [2185. Counting words with a given prefix.][2185]
Straight-forward, complexity `O(N)`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 3 - 93.10 | 0 - 100.00 | 0 - 100.00 | 39 - 96.80 |
| Mem Usage (MB-%)| 6.5 - 79.31 | 2.1 - 88.89 | 3.5 - 84.34 | 14 - 67.28 |
## [2194. Cells in a range on an excel sheet.][2194]
Find the differences between rows/cols and iterate through possible pair. Obligatory
`O(N^2)` solution.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 3 - 85.71 | 39 - 97.26 |
| Mem Usage (MB-%)| 6.7 - 42.86 | 2.2 - 46.15 | 3.1 - 87.76 | 13.9 - 21.52 |
## [2206. Divide array into equal pairs.][2206]
Use a set and continuously add/remove elements as you see them, i.e if its in, remove, if not
add it. By the end, if the set is empty, we have an equal number of pairs. Complexity `O(N)`.
C: Too bored to make the set, as usual.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| - | 0 - 100.00 | 7 - 95.56 | 97 - 90.09 |
| Mem Usage (MB-%)| - | 2.1 - 83.33 | 4.7 - 72.22 | 13.9 - 99.13 |
## [2220. Minimum bit flips to convert number.][2220]
Find smallest, count diff until smallest is zero using `&`, then count ones in largest (whatever remains of it).
Reduce by shifting right by `1`. Complexity is `O(d)` where `d` is the number of binary digits in the
largest number.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 31 - 95.09 |
| Mem Usage (MB-%)| 5.5 - 67.42 | 2 - 89.47 | 2 - 80.00 | 13.8 - 96.77 |
## [2235. Add two integers.][2235]
Yeah.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 61 - 14.21 |
| Mem Usage (MB-%)| 5.4 - 68.06 | 2.1 - 58.29 | 1.9 - 100.00 | 13.9 - 8.78 |
## [2236. Root Equals Sum Of Children.][2236]
Yeah, again.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 33 - 92.37 |
| Mem Usage (MB-%)| 5.6 - 74.30 | 2 - 69.01 | 1.9 - 94.48 | 13.8 - 94.80 |
## [2255. Count prefixes of a given string.][2255]
Pretty much exactly the same as [2185][2185].
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 6 - 96.88 | 2 - 100.00 | 3 - 93.44 | 53 - 98.74 |
| Mem Usage (MB-%)| 6.8 - 84.38 | 2.3 - 42.86 | 3.8 - 72.13 | 14.1 - 91.03 |
## [2278. Percentage of letter in string.][2278]
Relatively straight-forward, just need to do appropriate conversions for the strongly typed languages.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 36 - 84.79 |
| Mem Usage (MB-%)| 5.5 - 51.56 | 2 - 96.00 | 2 - 59.18 | 13.8 - 96.18 |
## [2315. Count Asterisks.][2315]
Keep a flag to denote when we are between two `'|'` and count the stars otherwise. `O(N)` where `N` is the
length of the string.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 32 - 94.05 |
| Mem Usage (MB-%)| 5.5 - 72.09 | 1.9 - 88.89 | 1.9 - 95.40 | 13.8 - 55.63 |
## [2319. Check if matrix is x matrix.][2319]
Traverse the matrix normally and simply check that positions do or don't have zeros. Worse case is `O(N^2)`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 52 - 95.08 | 0 - 100.00 | 31 - 100.00 | 284 - 94.06 |
| Mem Usage (MB-%)| 7.6 - 98.36 | 2.3 - 25.00 | 6.7 - 82.05 | 14.9 - 56.51 |
## [2325. Decode the message.][2325]
For languages that have it, use a map to create the mapping of characters to their secret character.
For C, use two arrays for a similar effect. For all cases, we iterate through both strings and perform
O(1) ops so the runtime is `O(N)`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 24 - 99.85 |
| Mem Usage (MB-%)| 5.9 - 89.66 | 2.1 - 58.33 | 2.6 - 69.29 | 13.8 - 82.69 |
## [2351. First letter to appear twice.][2351]
Use sets and return when the first check for membership is true. (I.e, set already contains
character). `O(N)`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 30 - 94.03 |
| Mem Usage (MB-%)| 5.4 - 99.43 | 2.2 - 30.49 | 1.9 - 41.18 | 13.8 - 94.46 |
## [2367. Smallest even multiple.][2367]
Loops through triples of nums, if at any point the difference is above the diff, we break.
`O(N^3)` solution despite that.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 38 - 97.00 |
| Mem Usage (MB-%)| 5.9 - 19.80 | 2.1 - 78.43 | 2.2 - 54.83 | 13.9 - 19.84 |
## [2389. Longest subsequence with limited sum.][2389]
Sort nums and create a new array of partial sums. Iterate through the queries and perform a
binary search on the partial sums array to find the index for which all elements in the partial
sums array are `<=` to the value of the query. Return that.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 29 - 95.74 | 0 - 100.00 | 16 - 72.99 | 123 - 90.30 |
| Mem Usage (MB-%)| 7.9 - 78.72 | 2.3 - 8.79 | 5.4 - 88.15 | 14.1 - 98.03 |
## [2399. Check distances between same letters.][2399]
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 40 - 97.49 |
| Mem Usage (MB-%)| 6.1 - 19.81 | 2 - 100.00 | 2.4 - 100.00 | 13.8 - 98.39 |
## [2413. Smallest even multiple.][2413]
Check if even/odd. If odd return `n * 2`, if even return `n`.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 32 - 100.00 |
| Mem Usage (MB-%)| 5.6 - 100.00 | 2.1 - 100.00 | 2 - 100.00 | 13.8 - 100.00 |
## [2418. Sort the people.][2418]
Sort `(name, height)` pair by height and return names. `O(NlogN)` due to sort.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 28 - 97.50 | 8 - 95.00 | 24 - 93.04 | 119 - 94.64 |
| Mem Usage (MB-%)| 10.8 - 57.50 | 2.3 - 90.83 | 6.8 - 92.41 | 14.3 - 89.83 |
## [2427. Number of common factors.][2427]
Go through values from `[1:(min(a, b)/2 + 1)]` and check if `a % i == 0 && b % i == 0`. If so, increment
the counter.
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 0 - 100.00 | 0 - 100.00 | 0 - 100.00 | 34 - 100.00 |
| Mem Usage (MB-%)| 5.7 - 100.00 | 2.1 - 100.00 | 1.9 - 100.00 | 13.9 - 66.67 |
## [2441. Largest positive integer that exists with its negative.][2441]
Sort (by absolute value) and go through pairwise looking for matching pairs, `O(NlogN)`. Another approach
is using a set and checking for membership of the negative of the current value, `O(N)` (but timings
didn't really back it up so most, except for Python, contain the sorting solution.)
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 22 - 82.61 | 4 - 72.97 | 25 - 88.41 | 125 - 98.65 |
| Mem Usage (MB-%)| 6.9 - 13.04 | 2.1 - 91.89 | 5.8 - 97.10 | 14.1 - 91.31 |
## [2455. Average value of even numbers that are divisible by three.][2455]
Sum and divide, straight-forward O(N).
| Stats/Lang | C | Rust | Go | Py |
|:-----------:|:--:|:-----:|:---:|:--:|
| Runtime (ms-%)| 8 - 90.59 | 0 - 100.00 | 3 - 100.00 | 74 - 99.30 |
| Mem Usage (MB-%)| 6.6 - 18.82 | 2.1 - 76.64 | 4.1 - 100.00 | 14 - 93.49 |
[1]: https://leetcode.com/problems/two-sum/
[2]: https://leetcode.com/problems/add-two-numbers/
[7]: https://leetcode.com/problems/reverse-integer/
[9]: https://leetcode.com/problems/palindrome-number/
[13]: https://leetcode.com/problems/roman-to-integer/
[14]: https://leetcode.com/problems/longest-common-prefix/
[21]: https://leetcode.com/problems/merge-two-sorted-lists
[24]: https://leetcode.com/problems/swap-nodes-in-pairs/
[27]: https://leetcode.com/problems/remove-element/
[28]: https://leetcode.com/problems/implement-strstr/
[31]: https://leetcode.com/problems/next-permutation/
[34]: https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/
[35]: https://leetcode.com/problems/search-insert-position/
[46]: https://leetcode.com/problems/permutations/
[50]: https://leetcode.com/problems/powx-n/
[53]: https://leetcode.com/problems/maximum-subarray
[58]: https://leetcode.com/problems/length-of-last-word/
[62]: https://leetcode.com/problems/unique-paths
[63]: https://leetcode.com/problems/unique-paths-ii/
[64]: https://leetcode.com/problems/summary-ranges
[66]: https://leetcode.com/problems/plus-one/
[69]: https://leetcode.com/problems/sqrtx/
[70]: https://leetcode.com/problems/climbing-stairs
[74]: https://leetcode.com/problems/search-a-2d-matrix/
[78]: https://leetcode.com/problems/subsets/
[83]: https://leetcode.com/problems/remove-duplicates-from-sorted-list/
[88]: https://leetcode.com/problems/merge-sorted-array/
[94]: https://leetcode.com/problems/binary-tree-inorder-traversal/
[100]: https://leetcode.com/problems/same-tree/
[101]: https://leetcode.com/problems/symmetric-tree/
[102]: https://leetcode.com/problems/binary-tree-level-order-traversal/
[104]: https://leetcode.com/problems/maximum-depth-of-binary-tree/
[107]: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
[108]: https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/
[110]: https://leetcode.com/problems/balanced-binary-tree/
[111]: https://leetcode.com/problems/minimum-depth-of-binary-tree/
[112]: https://leetcode.com/problems/path-sum/
[113]: https://leetcode.com/problems/path-sum-ii/
[116]: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
[118]: https://leetcode.com/problems/pascals-triangle/
[119]: https://leetcode.com/problems/pascals-triangle-ii/
[122]: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/
[125]: https://leetcode.com/problems/valid-palindrome/
[133]: https://leetcode.com/problems/clone-graph
[136]: https://leetcode.com/problems/single-number
[141]: https://leetcode.com/problems/linked-list-cycle/
[144]: https://leetcode.com/problems/binary-tree-preorder-traversal/
[145]: https://leetcode.com/problems/binary-tree-postorder-traversal/
[155]: https://leetcode.com/problems/min-stack/
[167]: https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
[171]: https://leetcode.com/problems/excel-sheet-column-number/
[190]: https://leetcode.com/problems/reverse-bits/
[191]: https://leetcode.com/problems/number-of-1-bits
[199]: https://leetcode.com/problems/binary-tree-right-side-view/
[201]: https://leetcode.com/problems/bitwise-and-of-numbers-range/
[206]: https://leetcode.com/problems/reverse-linked-list/
[217]: https://leetcode.com/problems/contains-duplicate/
[219]: https://leetcode.com/problems/contains-duplicate-ii
[220]: https://leetcode.com/problems/contains-duplicate-iii/
[225]: https://leetcode.com/problems/implement-stack-using-queues/
[226]: https://leetcode.com/problems/invert-binary-tree/
[231]: https://leetcode.com/problems/power-of-two/
[232]: https://leetcode.com/problems/implement-queue-using-stacks
[237]: https://leetcode.com/problems/delete-node-in-a-linked-list
[240]: https://leetcode.com/problems/search-a-2d-matrix-ii/
[242]: https://leetcode.com/problems/valid-anagram/
[257]: https://leetcode.com/problems/binary-tree-paths/
[258]: https://leetcode.com/problems/add-digits
[263]: https://leetcode.com/problems/ugly-number
[268]: https://leetcode.com/problems/missing-number/
[283]: https://leetcode.com/problems/move-zeroes
[292]: https://leetcode.com/problems/nim-game/
[326]: https://leetcode.com/problems/power-of-three/
[338]: https://leetcode.com/problems/counting-bits/
[342]: https://leetcode.com/problems/power-of-four/
[344]: https://leetcode.com/problems/reverse-string/
[345]: https://leetcode.com/problems/reverse-vowels-of-a-string/
[347]: https://leetcode.com/problems/top-k-frequent-elements/
[349]: https://leetcode.com/problems/intersection-of-two-arrays/
[367]: https://leetcode.com/problems/valid-perfect-square/
[374]: https://leetcode.com/problems/guess-number-higher-or-lower
[377]: https://leetcode.com/problems/combination-sum-iv/
[383]: https://leetcode.com/problems/ransom-note/
[384]: https://leetcode.com/problems/shuffle-an-array
[387]: https://leetcode.com/problems/first-unique-character-in-a-string/
[389]: https://leetcode.com/problems/find-the-difference/
[392]: https://leetcode.com/problems/is-subsequence/
[400]: https://leetcode.com/problems/nth-digit/
[404]: https://leetcode.com/problems/sum-of-left-leaves/
[405]: https://leetcode.com/problems/convert-a-number-to-hexadecimal/
[409]: https://leetcode.com/problems/longest-palindrome/
[412]: https://leetcode.com/problems/fizz-buzz/
[429]: https://leetcode.com/problems/n-ary-tree-level-order-traversal/
[434]: https://leetcode.com/problems/number-of-segments-in-a-string/
[448]: https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/
[451]: https://leetcode.com/problems/sort-characters-by-frequency
[461]: https://leetcode.com/problems/hamming-distance/
[476]: https://leetcode.com/problems/number-complement/
[485]: https://leetcode.com/problems/max-consecutive-ones/
[495]: https://leetcode.com/problems/teemo-attacking/
[496]: https://leetcode.com/problems/next-greater-element-
[500]: https://leetcode.com/problems/keyboard-row
[504]: https://leetcode.com/problems/base-7
[509]: https://leetcode.com/problems/fibonacci-number/
[515]: https://leetcode.com/problems/find-largest-value-in-each-tree-row/
[520]: https://leetcode.com/problems/detect-capital/
[521]: https://leetcode.com/problems/longest-uncommon-subsequence-i/
[530]: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
[535]: https://leetcode.com/problems/encode-and-decode-tinyurl/
[538]: https://leetcode.com/problems/convert-bst-to-greater-tree
[539]: https://leetcode.com/problems/minimum-time-difference/
[547]: https://leetcode.com/problems/number-of-provinces/
[551]: https://leetcode.com/problems/student-attendance-record-i/
[566]: https://leetcode.com/problems/reshape-the-matrix/
[572]: https://leetcode.com/problems/subtree-of-another-tree
[575]: https://leetcode.com/problems/distribute-candies
[589]: https://leetcode.com/problems/n-ary-tree-preorder-traversal
[559]: https://leetcode.com/problems/maximum-depth-of-n-ary-tree/
[561]: https://leetcode.com/problems/array-partition-i/
[590]: https://leetcode.com/problems/n-ary-tree-postorder-traversal
[605]: https://leetcode.com/problems/can-place-flowers/
[606]: https://leetcode.com/problems/construct-string-from-binary-tree/
[617]: https://leetcode.com/problems/merge-two-binary-trees/
[637]: https://leetcode.com/problems/average-of-levels-in-binary-tree/
[641]: https://leetcode.com/problems/design-circular-deque/
[643]: https://leetcode.com/problems/maximum-average-subarray-i/
[645]: https://leetcode.com/problems/set-mismatch/
[653]: https://leetcode.com/problems/two-sum-iv-input-is-a-bst/
[654]: https://leetcode.com/problems/maximum-binary-tree/
[657]: https://leetcode.com/problems/robot-return-to-origin/
[669]: https://leetcode.com/problems/trim-a-binary-search-tree/
[674]: https://leetcode.com/problems/longest-continuous-increasing-subsequence
[680]: https://leetcode.com/problems/valid-palindrome-ii/
[682]: https://leetcode.com/problems/baseball-game/
[684]: https://leetcode.com/problems/redundant-connection/
[690]: https://leetcode.com/problems/employee-importance
[693]: https://leetcode.com/problems/binary-number-with-alternating-bits/
[696]: https://leetcode.com/problems/count-binary-substrings/
[697]: https://leetcode.com/problems/degree-of-an-array/
[700]: https://leetcode.com/problems/search-in-a-binary-search-tree/
[704]: https://leetcode.com/problems/binary-search/
[705]: https://leetcode.com/problems/design-hashset
[709]: https://leetcode.com/problems/to-lower-case/
[728]: https://leetcode.com/problems/self-dividing-numbers
[744]: https://leetcode.com/problems/find-smallest-letter-greater-than-target/
[746]: https://leetcode.com/problems/min-cost-climbing-stairs/
[748]: https://leetcode.com/problems/shortest-completing-word/
[766]: https://leetcode.com/problems/toeplitz-matrix/
[771]: https://leetcode.com/problems/jewels-and-stones/
[783]: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
[788]: https://leetcode.com/problems/rotated-digits
[797]: https://leetcode.com/problems/all-paths-from-source-to-target/
[804]: https://leetcode.com/problems/unique-morse-code-words/
[806]: https://leetcode.com/problems/number-of-lines-to-write-string/
[807]: https://leetcode.com/problems/max-increase-to-keep-city-skyline/
[811]: https://leetcode.com/problems/subdomain-visit-count/
[812]: https://leetcode.com/problems/largest-triangle-area/
[821]: https://leetcode.com/problems/shortest-distance-to-a-character/
[824]: https://leetcode.com/problems/goat-latin/
[832]: https://leetcode.com/problems/flipping-an-image/
[841]: https://leetcode.com/problems/keys-and-rooms/
[845]: https://leetcode.com/problems/longest-mountain-in-array
[852]: https://leetcode.com/problems/peak-index-in-a-mountain-array/
[859]: https://leetcode.com/problems/buddy-strings/
[867]: https://leetcode.com/problems/transpose-matrix/
[868]: https://leetcode.com/problems/binary-gap/
[872]: https://leetcode.com/problems/leaf-similar-trees/
[876]: https://leetcode.com/problems/middle-of-the-linked-list/
[883]: https://leetcode.com/problems/projection-area-of-3d-shapes/
[884]: https://leetcode.com/problems/uncommon-words-from-two-sentences/
[888]: https://leetcode.com/problems/fair-candy-swap/
[896]: https://leetcode.com/problems/monotonic-array
[897]: https://leetcode.com/problems/increasing-order-search-tree/
[905]: https://leetcode.com/problems/sort-array-by-parity/
[908]: https://leetcode.com/problems/smallest-range-i/
[917]: https://leetcode.com/problems/reverse-only-letters/
[922]: https://leetcode.com/problems/sort-array-by-parity-ii/
[929]: https://leetcode.com/problems/unique-email-addresses/
[933]: https://leetcode.com/problems/number-of-recent-calls
[937]: https://leetcode.com/problems/reorder-data-in-log-files/
[938]: https://leetcode.com/problems/range-sum-of-bst/
[942]: https://leetcode.com/problems/di-string-match/
[944]: https://leetcode.com/problems/delete-columns-to-make-sorted
[961]: https://leetcode.com/problems/n-repeated-element-in-size-2n-array/
[965]: https://leetcode.com/problems/univalued-binary-tree/
[976]: https://leetcode.com/problems/largest-perimeter-triangle/
[977]: https://leetcode.com/problems/squares-of-a-sorted-array/
[985]: https://leetcode.com/problems/sum-of-even-numbers-after-queries
[989]: https://leetcode.com/problems/add-to-array-form-of-integer/
[997]: https://leetcode.com/problems/find-the-town-judge/
[999]: https://leetcode.com/problems/available-captures-for-rook
[1002]: https://leetcode.com/problems/find-common-characters/
[1008]: https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal
[1009]: https://leetcode.com/problems/complement-of-base-10-integer/
[1021]: https://leetcode.com/problems/remove-outermost-parentheses
[1022]: https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers
[1025]: https://leetcode.com/problems/divisor-game/
[1030]: https://leetcode.com/problems/matrix-cells-in-distance-order/
[1037]: https://leetcode.com/problems/valid-boomerang/
[1038]: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree
[1046]: https://leetcode.com/problems/last-stone-weight/
[1047]: https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/
[1051]: https://leetcode.com/problems/height-checker
[1071]: https://leetcode.com/problems/greatest-common-divisor-of-strings
[1078]: https://leetcode.com/problems/occurrences-after-bigram
[1103]: https://leetcode.com/problems/distribute-candies-to-people/
[1108]: https://leetcode.com/problems/defanging-an-ip-address/
[1122]: https://leetcode.com/problems/relative-sort-array/
[1128]: https://leetcode.com/problems/number-of-equivalent-domino-pairs/
[1137]: https://leetcode.com/problems/n-th-tribonacci-number/
[1146]: https://leetcode.com/problems/snapshot-array/
[1160]: https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/
[1170]: https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/
[1184]: https://leetcode.com/problems/distance-between-bus-stops/
[1185]: https://leetcode.com/problems/day-of-the-week/
[1189]: https://leetcode.com/problems/maximum-number-of-balloons/
[1200]: https://leetcode.com/problems/minimum-absolute-difference/
[1207]: https://leetcode.com/problems/unique-number-of-occurrences/
[1217]: https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/
[1221]: https://leetcode.com/problems/split-a-string-in-balanced-strings/
[1237]: https://leetcode.com/problems/find-positive-integer-solution-for-a-given-equation/
[1252]: https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/
[1260]: https://leetcode.com/problems/shift-2d-grid/
[1266]: https://leetcode.com/problems/minimum-time-visiting-all-points/
[1275]: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
[1281]: https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/
[1282]: https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to/
[1287]: https://leetcode.com/problems/element-appearing-more-than-25-in-sorted-array/
[1290]: https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/
[1295]: https://leetcode.com/problems/find-numbers-with-even-number-of-digits/
[1299]: https://leetcode.com/problems/replace-elements-with-greatest-element-on-right-side/
[1302]: https://leetcode.com/problems/deepest-leaves-sum/
[1304]: https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero
[1309]: https://leetcode.com/problems/decrypt-string-from-alphabet-to-integer-mapping
[1313]: https://leetcode.com/problems/decompress-run-length-encoded-list/
[1315]: https://leetcode.com/problems/sum-of-nodes-with-even-valued-grandparent/
[1317]: https://leetcode.com/problems/convert-integer-to-the-sum-of-two-no-zero-integers/
[1323]: https://leetcode.com/problems/maximum-69-number/
[1329]: https://leetcode.com/problems/sort-the-matrix-diagonally/
[1332]: https://leetcode.com/problems/remove-palindromic-subsequences/
[1337]: https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/
[1342]: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/
[1346]: https://leetcode.com/problems/check-if-n-and-its-double-exist/
[1351]: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix
[1356]: https://leetcode.com/problems/sort-integers-by-the-number-of-1-bits/
[1365]: https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
[1370]: https://leetcode.com/problems/increasing-decreasing-string/
[1374]: https://leetcode.com/problems/generate-a-string-with-characters-that-have-odd-counts/
[1379]: https://leetcode.com/problems/find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree/
[1380]: https://leetcode.com/problems/lucky-numbers-in-a-matrix/
[1381]: https://leetcode.com/problems/design-a-stack-with-increment-operation/
[1385]: https://leetcode.com/problems/find-the-distance-value-between-two-arrays
[1386]: https://leetcode.com/problems/sort-integers-by-the-power-value
[1389]: https://leetcode.com/problems/create-target-array-in-the-given-order/
[1394]: https://leetcode.com/problems/find-lucky-integer-in-an-array/
[1399]: https://leetcode.com/problems/count-largest-group/
[1403]: https://leetcode.com/problems/minimum-subsequence-in-non-increasing-order/
[1408]: https://leetcode.com/problems/string-matching-in-an-array/
[1409]: https://leetcode.com/problems/queries-on-a-permutation-with-key/
[1413]: https://leetcode.com/problems/minimum-value-to-get-positive-step-by-step-sum
[1414]: https://leetcode.com/problems/find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k/
[1417]: https://leetcode.com/problems/reformat-the-string/
[1422]: https://leetcode.com/problems/maximum-score-after-splitting-a-string/
[1431]: https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/
[1436]: https://leetcode.com/problems/destination-city/
[1441]: https://leetcode.com/problems/build-an-array-with-stack-operations/
[1446]: https://leetcode.com/problems/consecutive-characters
[1450]: https://leetcode.com/problems/number-of-students-doing-homework-at-a-given-time/
[1455]: https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/
[1460]: https://leetcode.com/problems/make-two-arrays-equal-by-reversing-sub-arrays/
[1464]: https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array
[1470]: https://leetcode.com/problems/shuffle-the-array/
[1475]: https://leetcode.com/problems/final-prices-with-a-special-discount-in-a-shop/
[1476]: https://leetcode.com/problems/subrectangle-queries/
[1480]: https://leetcode.com/problems/running-sum-of-1d-array/
[1486]: https://leetcode.com/problems/xor-operation-in-an-array/
[1491]: https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary
[1496]: https://leetcode.com/problems/path-crossing/
[1502]: https://leetcode.com/problems/can-make-arithmetic-progression-from-sequence/
[1507]: https://leetcode.com/problems/reformat-date/
[1512]: https://leetcode.com/problems/number-of-good-pairs/
[1518]: https://leetcode.com/problems/water-bottles/
[1523]: https://leetcode.com/problems/count-odd-numbers-in-an-interval-range/
[1528]: https://leetcode.com/problems/shuffle-string/
[1534]: https://leetcode.com/problems/count-good-triplets
[1539]: https://leetcode.com/problems/kth-missing-positive-number/
[1544]: https://leetcode.com/problems/make-the-string-great
[1550]: https://leetcode.com/problems/three-consecutive-odds
[1556]: https://leetcode.com/problems/thousand-separator/
[1557]: https://leetcode.com/problems/minimum-number-of-vertices-to-reach-all-nodes/
[1560]: https://leetcode.com/problems/most-visited-sector-in-a-circular-track/
[1572]: https://leetcode.com/problems/matrix-diagonal-sum/
[1582]: https://leetcode.com/problems/special-positions-in-a-binary-matrix
[1588]: https://leetcode.com/problems/sum-of-all-odd-length-subarrays/
[1598]: https://leetcode.com/problems/crawler-log-folder/
[1603]: https://leetcode.com/problems/design-parking-system/
[1608]: https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/
[1614]: https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses
[1619]: https://leetcode.com/problems/mean-of-array-after-removing-some-elements/
[1624]: https://leetcode.com/problems/largest-substring-between-two-equal-characters
[1629]: https://leetcode.com/problems/slowest-key/
[1636]: https://leetcode.com/problems/sort-array-by-increasing-frequency/
[1637]: https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/
[1640]: https://leetcode.com/problems/check-array-formation-through-concatenation
[1641]: https://leetcode.com/problems/count-sorted-vowel-strings
[1646]: https://leetcode.com/problems/get-maximum-in-generated-array/
[1652]: https://leetcode.com/problems/defuse-the-bomb/
[1656]: https://leetcode.com/problems/design-an-ordered-stream/
[1662]: https://leetcode.com/problems/check-if-two-string-arrays-are-equivalent
[1672]: https://leetcode.com/problems/richest-customer-wealth
[1678]: https://leetcode.com/problems/goal-parser-interpretation/
[1684]: https://leetcode.com/problems/count-the-number-of-consistent-strings/
[1688]: https://leetcode.com/problems/count-of-matches-in-tournament/
[1694]: https://leetcode.com/problems/reformat-phone-number
[1689]: https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/
[1700]: https://leetcode.com/problems/number-of-students-unable-to-eat-lunch/
[1704]: https://leetcode.com/problems/determine-if-string-halves-are-alike/
[1710]: https://leetcode.com/problems/maximum-units-on-a-truck/
[1716]: https://leetcode.com/problems/calculate-money-in-leetcode-bank/
[1720]: https://leetcode.com/problems/decode-xored-array/
[1725]: https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/
[1732]: https://leetcode.com/problems/find-the-highest-altitude
[1742]: https://leetcode.com/problems/maximum-number-of-balls-in-a-box/
[1748]: https://leetcode.com/problems/sum-of-unique-elements/
[1752]: https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/
[1768]: https://leetcode.com/problems/merge-strings-alternately/
[1773]: https://leetcode.com/problems/count-items-matching-a-rule
[1779]: https://leetcode.com/problems/find-nearest-point-that-has-the-same-x-or-y-coordinate/
[1790]: https://leetcode.com/problems/check-if-one-string-swap-can-make-strings-equal/
[1791]: https://leetcode.com/problems/find-center-of-star-graph/
[1796]: https://leetcode.com/problems/second-largest-digit-in-a-string/
[1797]: https://leetcode.com/problems/design-authentication-manager/
[1800]: https://leetcode.com/problems/maximum-ascending-subarray-sum
[1805]: https://leetcode.com/problems/number-of-different-integers-in-a-string/
[1812]: https://leetcode.com/problems/determine-color-of-a-chessboard-square/
[1816]: https://leetcode.com/problems/truncate-sentence/
[1822]: https://leetcode.com/problems/sign-of-the-product-of-an-array
[1827]: https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing
[1832]: https://leetcode.com/problems/check-if-the-sentence-is-pangram
[1837]: https://leetcode.com/problems/sum-of-digits-in-base-k/
[1844]: https://leetcode.com/problems/replace-all-digits-with-characters/
[1848]: https://leetcode.com/problems/minimum-distance-to-the-target-element/
[1854]: https://leetcode.com/problems/maximum-population-year/
[1859]: https://leetcode.com/problems/sorting-the-sentence
[1863]: https://leetcode.com/problems/sum-of-all-subset-xor-totals/
[1876]: https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/
[1880]: https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/
[1893]: https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered
[1897]: https://leetcode.com/problems/redistribute-characters-to-make-all-strings-equal/
[1902]: https://leetcode.com/problems/largest-odd-number-in-string/
[1908]: https://leetcode.com/problems/remove-one-element-to-make-the-array-strictly-increasing
[1912]: https://leetcode.com/problems/maximum-product-difference-between-two-pairs/
[1920]: https://leetcode.com/problems/build-array-from-permutation/
[1925]: https://leetcode.com/problems/count-square-sum-triples/
[1926]: https://leetcode.com/problems/nearest-exit-from-entrance-in-maze
[1929]: https://leetcode.com/problems/concatenation-of-array/
[1935]: https://leetcode.com/problems/maximum-number-of-words-you-can-type/
[1941]: https://leetcode.com/problems/check-if-all-characters-have-equal-number-of-occurrences
[1945]: https://leetcode.com/problems/sum-of-digits-of-string-after-convert/
[1961]: https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/
[1967]: https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/
[1974]: https://leetcode.com/problems/minimum-time-to-type-word-using-special-typewriter/
[1979]: https://leetcode.com/problems/find-greatest-common-divisor-of-array
[2000]: https://leetcode.com/problems/reverse-prefix-of-word/
[2006]: https://leetcode.com/problems/count-number-of-pairs-with-absolute-difference-k/
[2011]: https://leetcode.com/problems/final-value-of-variable-after-performing-operations/
[2037]: https://leetcode.com/problems/minimum-number-of-moves-to-seat-everyone/
[2042]: https://leetcode.com/problems/check-if-numbers-are-ascending-in-a-sentence/
[2057]: https://leetcode.com/problems/smallest-index-with-equal-value/
[2089]: https://leetcode.com/problems/find-target-indices-after-sorting-array/
[2103]: https://leetcode.com/problems/rings-and-rods/
[2114]: https://leetcode.com/problems/maximum-number-of-words-found-in-sentences/
[2119]: https://leetcode.com/problems/a-number-after-a-double-reversal/
[2124]: https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/
[2154]: https://leetcode.com/problems/keep-multiplying-found-values-by-two
[2160]: https://leetcode.com/problems/minimum-sum-of-four-digit-number-after-splitting-digits/
[2169]: https://leetcode.com/problems/count-operations-to-obtain-zero/
[2176]: https://leetcode.com/problems/count-equal-and-divisible-pairs-in-an-array
[2185]: https://leetcode.com/problems/counting-words-with-a-given-prefix/
[2194]: https://leetcode.com/problems/cells-in-a-range-on-an-excel-sheet
[2206]: https://leetcode.com/problems/divide-array-into-equal-pairs/
[2220]: https://leetcode.com/problems/minimum-bit-flips-to-convert-number
[2235]: https://leetcode.com/problems/add-two-integers
[2236]: https://leetcode.com/problems/root-equals-sum-of-children/
[2255]: https://leetcode.com/problems/count-prefixes-of-a-given-string
[2278]: https://leetcode.com/problems/percentage-of-letter-in-string/
[2315]: https://leetcode.com/problems/count-asterisks
[2319]: https://leetcode.com/problems/check-if-matrix-is-x-matrix
[2325]: https://leetcode.com/problems/decode-the-message
[2351]: https://leetcode.com/problems/first-letter-to-appear-twice
[2367]: https://leetcode.com/problems/number-of-arithmetic-triplets/
[2389]: https://leetcode.com/problems/longest-subsequence-with-limited-sum/
[2399]: https://leetcode.com/problems/check-distances-between-same-letters/
[2413]: https://leetcode.com/problems/smallest-even-multiple/
[2418]: https://leetcode.com/problems/sort-the-people
[2427]: https://leetcode.com/problems/number-of-common-factors
[2441]: https://leetcode.com/problems/largest-positive-integer-that-exists-with-its-negative/
[2455]: https://leetcode.com/problems/average-value-of-even-numbers-that-are-divisible-by-three/