Split array in three equal sum subarrays

• Difficulty Level : Medium
• Last Updated : 23 Nov, 2021

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.

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;    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;       // Loop until second last index    // as S2 should not be at the last      for (i = 0; i < n-1; 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 codeint main(){    int arr[] =  { 1, 3, 4, 0, 4 };    int n = sizeof(arr) / sizeof(arr);    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;        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;               // Loop until second last index        // as S2 should not be at the last        for (i = 0; i < n-1; 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    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     # Loop until second last index    # as S2 should not be at the last    for i in range(0,n-1):        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 codearr = [ 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;        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;             // Loop until second last index        // as S2 should not be at the last        for (i = 0; i < n-1; 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)



Javascript


Output
(1, 2)

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

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