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

Last Updated : 19 Oct, 2022

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 [4]

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 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]);         // 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

 # Python3 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 mp             = new Dictionary();         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

## Javascript



Output

4

Time Complexity: O(n) where n is the size of the given array
Auxiliary Space: O(n)

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[0]);       // Function call     cout << sumFirst(arr, n);       // This code is contributed by Arif }

## 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.*;   class GFG{       // Function to return the sum of // the first subarray public static 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 = Math.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 public static void main(String[] args) {     int arr[] = { 1, 3, 2, 1, 4 };     int n = arr.length;        // Function call     System.out.print(sumFirst(arr, n)); } }   // This code is contributed by divyeshrabadiya07

## Python3

 # Python3 program for Split the array into three # subarrays such that summation of first # and third subarray is equal and maximum import math   # Function to return the sum of # the first subarray def sumFirst(a, n):           # Two pointers are initialized     # one at the front and the other     # at the back     front_pointer = 0     back_pointer = n - 1           # prefixsum and suffixsum initialized     prefixsum = a[front_pointer]     suffixsum = a[back_pointer]           # answer variable initialized to 0     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 += 1             back_pointer -= 1               # prefixsum and suffixsum are updated             prefixsum += a[front_pointer]             suffixsum += a[back_pointer]                       elif (prefixsum > suffixsum):                           # If prefixsum is more,then back pointer is             # moved by one step and suffixsum updated.             back_pointer -= 1             suffixsum += a[back_pointer]         else:                           # If prefixsum is less,then front pointer is             # moved by one step and prefixsum updated.             front_pointer += 1             prefixsum += a[front_pointer]       # answer is returned     return answer   # Driver code arr = [ 1, 3, 2, 1, 4 ] n = len(arr)   # Function call print(sumFirst(arr, n))   # This code is contributed by Stream_Cipher

## C#

 // C# program for split the array into three // subarrays such that summation of first // and third subarray is equal and maximum using System;   class GFG{       // Function to return the sum of // the first subarray static 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 = Math.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 static void Main() {     int[] arr = { 1, 3, 2, 1, 4 };     int n = arr.Length;           // Function call     Console.WriteLine(sumFirst(arr, n)); } }   // This code is contributed by divyesh072019

## Javascript



Output

4

Complexity Analysis:

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

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