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https://github.com/seamoonpandey/probable-umbrella

Some coding stuffs.
https://github.com/seamoonpandey/probable-umbrella

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Some coding stuffs.

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# Discussion for the problems



## Contains Duplicate



**Intuition:**



The "Contains Duplicate" problem involves determining if a given list of integers contains any duplicate elements. Multiple solutions are available for this problem, each with its own trade-offs in terms of time and space complexity.



### Brute-Force Solution



One way to address the "Contains Duplicate" problem is to use a brute-force algorithm. This approach involves comparing every pair of elements in the list and has the following characteristics:



- **Time Complexity:** O(n^2) - It involves nested loops for element comparisons.

- **Space Complexity:** O(1) - No additional data structures are used.



**Sorting Solution:**



Another approach is to sort the array first and then compare corresponding indexes. This method offers better time complexity at the expense of space complexity:



- **Time Complexity:** O(nlogn) - Due to the sorting step.

- **Space Complexity:** O(1) - Minimal extra memory is used.



**Hash Table Solution:**



The third approach is to use a hash table, which sacrifices some space complexity for improved time complexity:



- **Time Complexity:** O(n) - Hash table lookup.

- **Space Complexity:** O(n) - Additional memory used for the hash table.



Now, let's add the discussion for the "Valid Anagram" problem you've provided:



## Valid Anagram



**Intuition:**



The "Valid Anagram" problem involves determining if any given two strings can be converted to each other by rearranging the alphabets.



**Sorting Solution:**



One approach to solving the "Valid Anagram" problem is to sort the strings using the `sorted(string)` method. This method offers a balance between time complexity and space complexity:



- **Time Complexity:** O(N*log(N)), where N is the length of the longer of the two input strings.

- **Space Complexity:** O(N), where N is the sum of the lengths of both strings.



**Hash Table Solution:**



To solve the "Valid Anagram" problem efficiently, you can use a hash table (dictionary) to keep track of the frequency of characters in both strings. Here's how the solution works:



1. Create two dictionaries, one for each input string (`s` and `t`), to store the frequency of characters.

2. Iterate through string `s` and increment the count of each character in the `s` dictionary.

3. Iterate through string `t` and decrement the count of each character in the `t` dictionary.

4. After both iterations, check if the dictionaries for `s` and `t` are equal. If they are, it means that the two strings can be rearranged to form each other, and you return `True`. Otherwise, return `False`.



- **Time Complexity:** Overall, the time complexity is O(N), where N is the length of the longer string.

- **Space Complexity:** The space complexity is determined by the dictionaries, which can contain at most all unique characters from both strings. In the worst case, if all characters are unique, the space complexity would be O(N), where N is the total length of both strings.