# Split array to three subarrays such that sum of first and third subarray is equal and maximum

Given an array of N integers, the task is to print the sum of the first subarray by splitting the array into exactly three subarrays such that the sum of the first and third subarray elements are equal and the maximum.

**Note:** All the elements must belong to a subarray and the subarrays can also be empty.

**Examples:**

Input:a[] = {1, 3, 1, 1, 4}

Output:5

Split the N numbers to [1, 3, 1], [] and [1, 4]

Input:a[] = {1, 3, 2, 1, 4}

Output:4

Split the N numbers to [1, 3], [2, 1] and [4]

A **naive approach** is to check for all possible partitions and use the prefix-sum concept to find out the partitions. The partition which gives the maximum sum of the first subarray will be the answer.

An **efficient approach** is as follows:

- Store the prefix sum and suffix sum of the N numbers.
- Hash the suffix sum’s index using a unordered_map in C++ or Hash-map in Java.
- Iterate from the beginning of the array, and check if the prefix sum exists in the suffix array beyond the current index i.
- If it does, then check for the previous maximum value and update accordingly.

Below is the implementation of the above approach:

`// C++ program for Split the array into three ` `// subarrays such that summation of first ` `// and third subarray is equal and maximum ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Fucntion to return the sum of ` `// the first subarray ` `int` `sumFirst(` `int` `a[], ` `int` `n) ` `{ ` ` ` `unordered_map<` `int` `, ` `int` `> mp; ` ` ` `int` `suf = 0; ` ` ` ` ` `// calculate the suffix sum ` ` ` `for` `(` `int` `i = n - 1; i >= 0; i--) { ` ` ` `suf += a[i]; ` ` ` `mp[suf] = i; ` ` ` `} ` ` ` ` ` `int` `pre = 0; ` ` ` `int` `maxi = -1; ` ` ` ` ` `// iterate from beginning ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// prefix sum ` ` ` `pre += a[i]; ` ` ` ` ` `// check if it exists beyond i ` ` ` `if` `(mp[pre] > i) { ` ` ` ` ` `// if greater then previous ` ` ` `// then update maximum ` ` ` `if` `(pre > maxi) { ` ` ` `maxi = pre; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// First and second subarray empty ` ` ` `if` `(maxi == -1) ` ` ` `return` `0; ` ` ` ` ` `// partition done ` ` ` `else` ` ` `return` `maxi; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `a[] = { 1, 3, 2, 1, 4 }; ` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `cout << sumFirst(a, n); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

4

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