# Split array in three equal sum subarrays

Consider an array A of n integers. Determine if array A can be split into three consecutive parts such that sum of each part is equal. If yes then print any index pair(i, j) such that sum(arr[0..i]) = sum(arr[i+1..j]) = sum(arr[j+1..n-1]), otherwise print -1.

Examples:

```Input : arr[] = {1, 3, 4, 0, 4}
Output : (1, 2)
Sum of subarray arr[0..1] is equal to
sum of subarray arr[2..3] and also to
sum of subarray arr[4..4]. The sum is 4.

Input : arr[] = {2, 3, 4}
Output : -1
No three subarrays exist which have equal
sum.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to first find all the subarrays, store the sum of these subarrays with their starting and ending points, and then find three disjoint continuous subarrays with equal sum. Time complexity of this solution will be quadratic.

An efficient solution is to first find the sum S of all array elements. Check if this sum is divisible by 3 or not. This is because if sum is not divisible then the sum cannot be split in three equal sum sets. If there are three contiguous subarrays with equal sum, then sum of each subarray is S/3. Suppose the required pair of indices is (i, j) such that sum(arr[0..i]) = sum(arr[i+1..j]) = S/3. Also sum(arr[0..i]) = preSum[i] and sum(arr[i+1..j]) = preSum[j] – preSum[i]. This gives preSum[i] = preSum[j] – preSum[i] = S/3. This gives preSum[j] = 2*preSum[i]. Thus, the problem reduces to find two indices i and j such that preSum[i] = S/3 and preSum[j] = 2*(S/3).
For finding these two indices, traverse the array and store sum upto current element in a variable preSum. Check if preSum is equal to S/3 and 2*(S/3).

Implementation:

## C++

 `// CPP program to determine if array arr[] ` `// can be split into three equal sum sets. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to determine if array arr[] ` `// can be split into three equal sum sets. ` `int` `findSplit(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `i; ` ` `  `    ``// variable to store prefix sum ` `    ``int` `preSum = 0; ` ` `  `    ``// variables to store indices which  ` `    ``// have prefix sum divisible by S/3. ` `    ``int` `ind1 = -1, ind2 = -1; ` ` `  `    ``// variable to store sum of ` `    ``// entire array. ` `    ``int` `S; ` ` `  `    ``// Find entire sum of the array. ` `    ``S = arr[0]; ` `    ``for` `(i = 1; i < n; i++)  ` `        ``S += arr[i]; ` ` `  `    ``// Check if array can be split in ` `    ``// three equal sum sets or not. ` `    ``if``(S % 3 != 0) ` `        ``return` `0; ` `     `  `    ``// Variables to store sum S/3  ` `    ``// and 2*(S/3). ` `    ``int` `S1 = S / 3; ` `    ``int` `S2 = 2 * S1; ` ` `  `    ``for` `(i = 0; i < n; i++) ` `    ``{ ` `        ``preSum += arr[i]; ` `         `  `        ``// If prefix sum is equal to S/3 ` `        ``// store current index. ` `        ``if` `(preSum == S1 && ind1 == -1) ` `            ``ind1 = i; ` `         `  `        ``// If prefix sum is equal to 2* (S/3) ` `        ``// store current index. ` `        ``else` `if``(preSum  == S2 && ind1 != -1) ` `        ``{ ` `            ``ind2 = i; ` `             `  `            ``// Come out of the loop as both the ` `            ``// required indices are found. ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// If both the indices are found  ` `    ``// then print them. ` `    ``if` `(ind1 != -1 && ind2 != -1) ` `    ``{ ` `        ``cout << ``"("` `<< ind1 << ``", "` `                                ``<< ind2 << ``")"``; ` `        ``return` `1; ` `    ``} ` ` `  `    ``// If indices are not found return 0. ` `    ``return` `0; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] =  { 1, 3, 4, 0, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``if` `(findSplit(arr, n) == 0)  ` `        ``cout << ``"-1"``;  ` `    ``return` `0; ` `} `

## Java

 `// Java program to determine if array arr[] ` `// can be split into three equal sum sets. ` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `public` `class` `GFG { ` `     `  `    ``// Function to determine if array arr[] ` `    ``// can be split into three equal sum sets. ` `    ``static` `int` `findSplit(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `i; ` `     `  `        ``// variable to store prefix sum ` `        ``int` `preSum = ``0``; ` `     `  `        ``// variables to store indices which  ` `        ``// have prefix sum divisible by S/3. ` `        ``int` `ind1 = -``1``, ind2 = -``1``; ` `     `  `        ``// variable to store sum of ` `        ``// entire array. ` `        ``int` `S; ` `     `  `        ``// Find entire sum of the array. ` `        ``S = arr[``0``]; ` `        ``for` `(i = ``1``; i < n; i++)  ` `            ``S += arr[i]; ` `     `  `        ``// Check if array can be split in ` `        ``// three equal sum sets or not. ` `        ``if``(S % ``3` `!= ``0``) ` `            ``return` `0``; ` `         `  `        ``// Variables to store sum S/3  ` `        ``// and 2*(S/3). ` `        ``int` `S1 = S / ``3``; ` `        ``int` `S2 = ``2` `* S1; ` `     `  `        ``for` `(i = ``0``; i < n; i++) ` `        ``{ ` `            ``preSum += arr[i]; ` `             `  `        ``// If prefix sum is equal to S/3 ` `        ``// store current index. ` `            ``if` `(preSum == S1 && ind1 == -``1``) ` `                ``ind1 = i; ` `             `  `        ``// If prefix sum is equal to 2*(S/3) ` `        ``// store current index. ` `            ``else` `if``(preSum == S2 && ind1 != -``1``) ` `            ``{ ` `                ``ind2 = i; ` `                 `  `                ``// Come out of the loop as both the ` `                ``// required indices are found. ` `                ``break``; ` `            ``} ` `        ``} ` `     `  `        ``// If both the indices are found  ` `        ``// then print them. ` `        ``if` `(ind1 != -``1` `&& ind2 != -``1``) ` `        ``{ ` `            ``System.out.print(``"("` `+ ind1 + ``", "` `                            ``+ ind2 + ``")"``); ` `            ``return` `1``; ` `        ``} ` `     `  `        ``// If indices are not found return 0. ` `        ``return` `0``; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `[]arr = { ``1``, ``3``, ``4``, ``0``, ``4` `}; ` `        ``int` `n = arr.length; ` `        ``if` `(findSplit(arr, n) == ``0``)  ` `            ``System.out.print(``"-1"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Manish Shaw  ` `// (manishshaw1) `

## Python3

 `# Python3 program to determine if array arr[] ` `# can be split into three equal sum sets. ` ` `  `# Function to determine if array arr[] ` `# can be split into three equal sum sets. ` `def` `findSplit(arr, n): ` `    ``# variable to store prefix sum ` `    ``preSum ``=` `0` ` `  `    ``# variables to store indices which  ` `    ``# have prefix sum divisible by S/3. ` `    ``ind1 ``=` `-``1`  `    ``ind2 ``=` `-``1` ` `  `    ``# variable to store sum of ` `    ``# entire array. S ` ` `  `    ``# Find entire sum of the array. ` `    ``S ``=` `arr[``0``] ` `    ``for` `i ``in` `range``(``1``, n): ` `        ``S ``+``=` `arr[i] ` ` `  `    ``# Check if array can be split in ` `    ``# three equal sum sets or not. ` `    ``if``(S ``%` `3` `!``=` `0``): ` `        ``return` `0` `     `  `    ``# Variables to store sum S/3  ` `    ``# and 2*(S/3). ` `    ``S1 ``=` `S ``/` `3` `    ``S2 ``=` `2` `*` `S1 ` ` `  `    ``for` `i ``in` `range``(``0``,n): ` `        ``preSum ``+``=` `arr[i] ` `         `  `        ``# If prefix sum is equal to S/3 ` `        ``# store current index. ` `        ``if` `(preSum ``=``=` `S1 ``and` `ind1 ``=``=` `-``1``): ` `            ``ind1 ``=` `i ` `        ``# If prefix sum is equal to 2*(S/3) ` `        ``# store current index.         ` `        ``elif``(preSum ``=``=` `S2 ``and` `ind1 !``=` `-``1``): ` `            ``ind2 ``=` `i ` `             `  `            ``# Come out of the loop as both the ` `            ``# required indices are found. ` `            ``break`     ` `  `    ``# If both the indices are found  ` `    ``# then print them. ` `    ``if` `(ind1 !``=` `-``1` `and` `ind2 !``=` `-``1``): ` `        ``print` `(``"({}, {})"``.``format``(ind1,ind2)) ` `        ``return` `1` `     `  `    ``# If indices are not found return 0. ` `    ``return` `0` ` `  `# Driver code ` `arr ``=` `[ ``1``, ``3``, ``4``, ``0``, ``4` `] ` `n ``=` `len``(arr) ` `if` `(findSplit(arr, n) ``=``=` `0``) : ` `    ``print` `(``"-1"``)  ` `# This code is contributed by Manish Shaw ` `# (manishshaw1) `

## C#

 `// C# program to determine if array arr[] ` `// can be split into three equal sum sets. ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG { ` `     `  `    ``// Function to determine if array arr[] ` `    ``// can be split into three equal sum sets. ` `    ``static` `int` `findSplit(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `i; ` `     `  `        ``// variable to store prefix sum ` `        ``int` `preSum = 0; ` `     `  `        ``// variables to store indices which  ` `        ``// have prefix sum divisible by S/3. ` `        ``int` `ind1 = -1, ind2 = -1; ` `     `  `        ``// variable to store sum of ` `        ``// entire array. ` `        ``int` `S; ` `     `  `        ``// Find entire sum of the array. ` `        ``S = arr[0]; ` `        ``for` `(i = 1; i < n; i++)  ` `            ``S += arr[i]; ` `     `  `        ``// Check if array can be split in ` `        ``// three equal sum sets or not. ` `        ``if``(S % 3 != 0) ` `            ``return` `0; ` `         `  `        ``// Variables to store sum S/3  ` `        ``// and 2*(S/3). ` `        ``int` `S1 = S / 3; ` `        ``int` `S2 = 2 * S1; ` `     `  `        ``for` `(i = 0; i < n; i++) ` `        ``{ ` `            ``preSum += arr[i]; ` `             `  `        ``// If prefix sum is equal to S/3 ` `        ``// store current index. ` `            ``if` `(preSum ==  S1 && ind1 == -1) ` `                ``ind1 = i; ` `             `  `        ``// If prefix sum is equal to S/3 ` `        ``// store current index. ` `            ``else` `if``(preSum == S2 && ind1 != -1) ` `            ``{ ` `                ``ind2 = i; ` `                 `  `                ``// Come out of the loop as both the ` `                ``// required indices are found. ` `                ``break``; ` `            ``} ` `        ``} ` `     `  `        ``// If both the indices are found  ` `        ``// then print them. ` `        ``if` `(ind1 != -1 && ind2 != -1) ` `        ``{ ` `            ``Console.Write(``"("` `+ ind1 + ``", "`  `                             ``+ ind2 + ``")"``); ` `            ``return` `1; ` `        ``} ` `     `  `        ``// If indices are not found return 0. ` `        ``return` `0; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = { 1, 3, 4, 0, 4 }; ` `        ``int` `n = arr.Length; ` `        ``if` `(findSplit(arr, n) == 0)  ` `            ``Console.Write(``"-1"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Manish Shaw  ` `// (manishshaw1) `

## PHP

 ` `

Output:

```(1, 2)
```

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

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