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https://github.com/nxhhoang/maximum-subarray-sum-3-in-marisaoj

Solution developing
https://github.com/nxhhoang/maximum-subarray-sum-3-in-marisaoj

competitive-programming

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Solution developing

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# Maximum-subarray-sum-3-in-marisaoj



The solution I found for this problem can be used for all k (the numbers of non-overlapping, non-empty subarrays) which are smaller than n (the size of array).

The time complexity of this solution is **O(n * k)**.



The solution comes up when I tried to solve maximum-subarray-sum-1 and maximum-subarray-sum-2 using prefix Sums.



# Maximum-subarray-sum-1

This can be solved using $O(n)$ time complexity one. 



The idea is to find the maximum sums from l to r with $1 <= l < r <= n$. This sum is equal to $S_r - S_l$ with $1 <= l < r <= n$, this leads to the need of finding the minimum value of $S_l$. That means for every $r$ $(1 <= r <= n)$ 

we need to find the smallest prefix sum with the length of $l$ $(l < r)$. This can be solved using one loop.



$ans_r = prefix_r - min(prefix_l)$ with $1 <= l < r <= n$.



If we use a loop to run $r$ through 1 to $n$, we can update min rolling each $r$; This allows us to calculate the minimum value with O(1) complexity. The answer to the problem is the maximum value of $S_r - S_l$ with $1<=r<=n$.



The solution can be found in other websites.



# Maximum-subarray-sum-2

The requirement of this problem is to find $1 <= l <= r < i <= j <= n$ such that $B = (A_l + A_{l + 1} + ... + A_r) + (A_i + A_{i + 1} + ... + A_j)$ is maximized

With the idea of **Maximum-subarray-sum-1**, we change this to an equivalent expression.



**Sol_1**



Consider $S_k = A_1 + A_2 + ... + A_k$



$B = S_j - (A_{i - 1} + ... + A_{r + 1}) - (A_{l - 1} + A_{l - 2} + ... + A_1)$



  $= S_j - (S_{i - 1} - (A_r + A_{r - 1} + ... + A_l))$

  

  $= S_j - (S_{i - 1} - (S_r - S_{l - 1}))$



So to find the maximum of $B$, we need to find the minimum value of this $(S_{i - 1} - (S_r - S_{l - 1}))$, which can be found when the value of $(S_r - S_{l - 1})$ is maximized



Check __max-sub-sum-2-ver1.cpp__ to see the executed solution.



**Sol_2** This problem can also be solved when loop through $h$ from 1 to $n$ and find the maximum value of a sum from $1$ to $h$ and a sum from $h$ to $n$ $(1 <= l <= r <= h < i <= j <= n)$.



# Maximum-subarray-sum-3

Still with the idea from above problems, with $1 <= l <= r < i <= j < p <= k <= n$ such that $M = (A_l + A_{l + 1} + ... + A_r) + (A_i + A_{i + 1} + ... + A_j) + (A_p + A_{p + 1} + ... + A_k)$ is maximized



Check **max-sub-sum-3-ver1.cpp** (this solution is the combination of **Sol_1** and **Sol_2** from **Maximum-subarray-sum-2**) 



We used the **B** expression from _maximum-subarray-sum-2_.



$M = (A_l + A_{l + 1} + ... + A_r) + (A_i + A_{i + 1} + ... + A_j) + (A_p + A_{p + 1} + ... + A_k)$



  $= B + (A_p + A_{p + 1} + ... + A_k)$

  

  $= S_p - S_{k - 1} + B$

  

  $= S_p - (S_{k - 1} - B)$



Similarly, to find the maximum of M, we need to find the minimum value of this $(S_{k - 1} - B)$, which can be found when the value of $B$ is maximized



Check __max-sub-sum-3-ver2.cpp__ to see the answer.



A little optimization for memory, check __max-sub-sum-3-ver3.cpp__ 



Continue optimizing - shorten the code, check __max-sub-sum-3-ver4.cpp__.

This solution beats all solutions from other people in terms of time complexity and space complexity



![image](https://github.com/user-attachments/assets/c71ac419-2c44-4c8f-8b11-4424b460678e)



**This has led to a generalized solution for any k <= n, check __max-sub-sum-3-ver5.cpp__, but remember to change k to the number you want**



With this code, I chose k = 2 for **maximum-subarray-sum-2**, check this code __max-sub-sum-2-ver2.cpp__ and the result can be shown in this image



![image](https://github.com/user-attachments/assets/3a301569-62ef-4924-bf61-9c3c7dadeb81)



Thank you for reading!



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