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

Input: a[] = {1, 3, 2, 1, 4}
Output:
Split the N numbers to [1, 3], [2, 1] and 

METHOD 1
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++

 `// C++ program for Split the array into three` `// subarrays such that summation of first` `// and third subarray is equal and maximum` `#include ` `using` `namespace` `std;`   `// Function 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);` `  `  `    ``// Function call` `    ``cout << sumFirst(a, n);` `    ``return` `0;` `}`

## Java

 `// Java program for Split the array into three` `// subarrays such that summation of first` `// and third subarray is equal and maximum` `import` `java.util.HashMap;` `import` `java.util.Map;`   `class` `GfG {`   `    ``// Function to return the sum` `    ``// of the first subarray` `    ``static` `int` `sumFirst(``int` `a[], ``int` `n)` `    ``{` `        ``HashMap mp = ``new` `HashMap<>();` `        ``int` `suf = ``0``;`   `        ``// calculate the suffix sum` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)` `        ``{` `            ``suf += a[i];` `            ``mp.put(suf, i);` `        ``}`   `        ``int` `pre = ``0``, 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.containsKey(pre) && mp.get(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` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `a[] = { ``1``, ``3``, ``2``, ``1``, ``4` `};` `        ``int` `n = a.length;` `      `  `        ``// Function call` `        ``System.out.println(sumFirst(a, n));` `    ``}` `}`   `// This code is contributed by Rituraj Jain`

## Python3

 `# Python 3 program for Split the array into three` `# subarrays such that summation of first` `# and third subarray is equal and maximum`   `# Function to return the sum of` `# the first subarray`     `def` `sumFirst(a, n):` `    ``mp ``=` `{i: ``0` `for` `i ``in` `range``(``7``)}` `    ``suf ``=` `0` `    ``i ``=` `n ``-` `1`   `    ``# calculate the suffix sum` `    ``while``(i >``=` `0``):` `        ``suf ``+``=` `a[i]` `        ``mp[suf] ``=` `i` `        ``i ``-``=` `1`   `    ``pre ``=` `0` `    ``maxi ``=` `-``1`   `    ``# iterate from beginning` `    ``for` `i ``in` `range``(n):`   `        ``# 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` `if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `[``1``, ``3``, ``2``, ``1``, ``4``]` `    ``n ``=` `len``(a)` `    `  `    ``# Function call` `    ``print``(sumFirst(a, n))`   `# This code is contributed by` `# Surendra_Gangwar`

## C#

 `// C# program for Split the array into three` `// subarrays such that summation of first` `// and third subarray is equal and maximum` `using` `System;` `using` `System.Collections.Generic;`   `class` `GfG {`   `    ``// Function to return the sum` `    ``// of the first subarray` `    ``static` `int` `sumFirst(``int``[] a, ``int` `n)` `    ``{` `        ``Dictionary<``int``, ``int``> mp` `            ``= ``new` `Dictionary<``int``, ``int``>();` `        ``int` `suf = 0;`   `        ``// calculate the suffix sum` `        ``for` `(``int` `i = n - 1; i >= 0; i--) ` `        ``{` `            ``suf += a[i];` `            ``mp.Add(suf, i);` `            ``if` `(mp.ContainsKey(suf))` `            ``{` `                ``mp.Remove(suf);` `            ``}` `            ``mp.Add(suf, i);` `        ``}`   `        ``int` `pre = 0, 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.ContainsKey(pre) && 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` `    ``public` `static` `void` `Main(String[] args)` `    ``{`   `        ``int``[] a = { 1, 3, 2, 1, 4 };` `        ``int` `n = a.Length;` `      `  `        ``// Function call` `        ``Console.WriteLine(sumFirst(a, n));` `    ``}` `}`   `// This code is contributed by Rajput-Ji`

Output

```4
```

METHOD 2

Approach: We will use two pointers concept where one pointer will start from the front and the other from the back. In each iteration, the sum of the first and last subarray is compared and if they are the same then the sum is updated in the answer variable.

Algorithm:

• Initialize front_pointer to 0 and back_pointer to n-1.
• Initialize prefixsum to arr[ front_pointer ] and suffixsum to arr[back_pointer].
• The summations are compared.
If prefixsum > suffixsum ,back_pointer is decremented by 1 and suffixsum+= arr[ back_pointer ].
If prefixsum < suffixsum, front_pointer is incremented by 1 and prefixsum+= arr[ front_pointer ]
If they are the same then the sum is updated in the answer variable and both the pointers are moved by one step
and both prefixsum and suffixsum are updated accordingly.
• The above step is continued until the front_pointer is no less than the back_pointer.

Below is the implementation of the above approach:

## C++

 `// C++ program for Split the array into three` `// subarrays such that summation of first` `// and third subarray is equal and maximum` `#include ` `using` `namespace` `std;`   `// Function to return the sum of` `// the first subarray` `int` `sumFirst(``int` `a[], ``int` `n)` `{` `    ``// two pointers are initialized` `    ``// one at the front and the other` `    ``// at the back` `    ``int` `front_pointer = 0;` `    ``int` `back_pointer = n - 1;`   `    ``// prefixsum and suffixsum initialized` `    ``int` `prefixsum = a[front_pointer];` `    ``int` `suffixsum = a[back_pointer];`   `    ``// answer variable initialized to 0` `    ``int` `answer = 0;`   `    ``while` `(front_pointer < back_pointer) ` `    ``{` `        ``// if the summation are equal` `        ``if` `(prefixsum == suffixsum)` `        ``{` `            ``// answer updated` `            ``answer = max(answer, prefixsum);`   `            ``// both the pointers are moved by step` `            ``front_pointer++;` `            ``back_pointer--;`   `            ``// prefixsum and suffixsum are updated` `            ``prefixsum += a[front_pointer];` `            ``suffixsum += a[back_pointer];` `        ``}` `        ``else` `if` `(prefixsum > suffixsum)` `        ``{` `            ``// if prefixsum is more,then back pointer is` `            ``// moved by one step and suffixsum updated.` `            ``back_pointer--;` `            ``suffixsum += a[back_pointer];` `        ``}` `        ``else` `        ``{` `            ``// if prefixsum is less,then front pointer is` `            ``// moved by one step and prefixsum updated.` `            ``front_pointer++;` `            ``prefixsum += a[front_pointer];` `        ``}` `    ``}`   `    ``// answer is returned` `    ``return` `answer;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1, 3, 2, 1, 4 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Function call` `    ``cout << sumFirst(arr, n);`   `    ``// This code is contributed by Arif` `}`

Output

```4
```

Time Complexity: O(n)
Auxiliary Space: O(1)

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