# Print indices of pair of array elements required to be removed to split array into 3 equal sum subarrays

• Last Updated : 22 Nov, 2021

Given an array arr[] consisting of N integers, the task is to print the indices of two array elements required to be removed such that the given array can be split into three subarrays of equal sum. If not possible to do so, then print “-1”.

Examples:

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Input: arr[] = {2, 5, 12, 7, 19, 4, 3}
Output: 2 4
Explanation:
Removing arr[2] and arr[4] modifies arr[] to {2, 5, 7, 4, 3}.
Sum of subarray {arr[0], arr[1]} = 7.
arr[2] = 7.
Sum of subarray {arr[3], arr[4]} = 7.

Input: arr[] = {2, 1, 13, 5, 14}
Output: -1

Naive Approach: The simplest approach is to generate all possible pairs of array elements and for each pair, check if removal of these pairs can generate three equal sum subarrays from the given array.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to check if array can``// be split into three equal sum``// subarrays by removing two elements``void` `findSplit(``int` `arr[], ``int` `N)``{``    ``for` `(``int` `l = 1; l <= N - 4; l++) {` `        ``for` `(``int` `r = l + 2; r <= N - 2; r++) {` `            ``// Stores sum of all three subarrays``            ``int` `lsum = 0, rsum = 0, msum = 0;` `            ``// Sum of left subarray``            ``for` `(``int` `i = 0; i <= l - 1; i++) {``                ``lsum += arr[i];``            ``}` `            ``// Sum of middle subarray``            ``for` `(``int` `i = l + 1; i <= r - 1; i++) {``                ``msum += arr[i];``            ``}` `            ``// Sum of right subarray``            ``for` `(``int` `i = r + 1; i < N; i++) {``                ``rsum += arr[i];``            ``}` `            ``// Check if sum of subarrays are equal``            ``if` `(lsum == rsum && rsum == msum) {` `                ``// Print the possible pair``                ``cout << l << ``" "` `<< r << endl;``                ``return``;``            ``}``        ``}``    ``}` `    ``// If no pair exists, print -1``    ``cout << -1 << endl;``}` `// Driver code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 2, 5, 12, 7, 19, 4, 3 };` `    ``// Size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``findSplit(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `// Function to check if array can``// be split into three equal sum``// subarrays by removing two elements``static` `void` `findSplit(``int` `arr[], ``int` `N)``{``    ``for` `(``int` `l = ``1``; l <= N - ``4``; l++)``    ``{``        ``for` `(``int` `r = l + ``2``; r <= N - ``2``; r++)``        ``{` `            ``// Stores sum of all three subarrays``            ``int` `lsum = ``0``, rsum = ``0``, msum = ``0``;` `            ``// Sum of left subarray``            ``for` `(``int` `i = ``0``; i <= l - ``1``; i++) {``                ``lsum += arr[i];``            ``}` `            ``// Sum of middle subarray``            ``for` `(``int` `i = l + ``1``; i <= r - ``1``; i++) {``                ``msum += arr[i];``            ``}` `            ``// Sum of right subarray``            ``for` `(``int` `i = r + ``1``; i < N; i++) {``                ``rsum += arr[i];``            ``}` `            ``// Check if sum of subarrays are equal``            ``if` `(lsum == rsum && rsum == msum) {` `                ``// Print the possible pair``                ``System.out.println( l + ``" "` `+ r );``                ``return``;``            ``}``        ``}``    ``}` `    ``// If no pair exists, print -1``    ``System.out.print(-``1` `);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ` `    ``// Given array``    ``int` `arr[] = { ``2``, ``5``, ``12``, ``7``, ``19``, ``4``, ``3` `};` `    ``// Size of the array``    ``int` `N = arr.length;``    ``findSplit(arr, N);``}``}` `// This code is contributed by sanjoy_62.`

## Python3

 `# Pyhton 3 program for the above approach` `# Function to check if array can``# be split into three equal sum``# subarrays by removing two elements``def` `findSplit(arr, N):``    ``for` `l ``in` `range``(``1``, N ``-` `3``, ``1``):``        ``for` `r ``in` `range``(l ``+` `2``, N ``-` `1``, ``1``):``          ` `            ``# Stores sum of all three subarrays``            ``lsum ``=` `0``            ``rsum ``=` `0``            ``msum ``=` `0` `            ``# Sum of left subarray``            ``for` `i ``in` `range``(``0``, l, ``1``):``                ``lsum ``+``=` `arr[i]` `            ``# Sum of middle subarray``            ``for` `i ``in` `range``(l ``+` `1``, r, ``1``):``                ``msum ``+``=` `arr[i]` `            ``# Sum of right subarray``            ``for` `i ``in` `range``(r ``+` `1``, N, ``1``):``                ``rsum ``+``=` `arr[i]` `            ``# Check if sum of subarrays are equal``            ``if` `(lsum ``=``=` `rsum ``and` `rsum ``=``=` `msum):``              ` `                ``# Print the possible pair``                ``print``(l, r)``                ``return` `    ``# If no pair exists, print -1``    ``print``(``-``1``)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given array``    ``arr ``=`  `[``2``, ``5``, ``12``, ``7``, ``19``, ``4``, ``3``]``    ` `    ``# Size of the array``    ``N ``=` `len``(arr)``    ``findSplit(arr, N)``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{` `  ``// Function to check if array can``  ``// be split into three equal sum``  ``// subarrays by removing two elements``  ``static` `void` `findSplit(``int` `[]arr, ``int` `N)``  ``{``    ``for` `(``int` `l = 1; l <= N - 4; l++)``    ``{``      ``for` `(``int` `r = l + 2; r <= N - 2; r++)``      ``{` `        ``// Stores sum of all three subarrays``        ``int` `lsum = 0, rsum = 0, msum = 0;` `        ``// Sum of left subarray``        ``for` `(``int` `i = 0; i <= l - 1; i++) {``          ``lsum += arr[i];``        ``}` `        ``// Sum of middle subarray``        ``for` `(``int` `i = l + 1; i <= r - 1; i++) {``          ``msum += arr[i];``        ``}` `        ``// Sum of right subarray``        ``for` `(``int` `i = r + 1; i < N; i++) {``          ``rsum += arr[i];``        ``}` `        ``// Check if sum of subarrays are equal``        ``if` `(lsum == rsum && rsum == msum) {` `          ``// Print the possible pair``          ``Console.WriteLine( l + ``" "` `+ r );``          ``return``;``        ``}``      ``}``    ``}` `    ``// If no pair exists, print -1``    ``Console.Write(-1 );``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{` `    ``// Given array``    ``int` `[]arr = { 2, 5, 12, 7, 19, 4, 3 };` `    ``// Size of the array``    ``int` `N = arr.Length;``    ``findSplit(arr, N);``  ``}``}` `// This code is contributed by AnkThon`

## Javascript

 ``
Output:
`2 4`

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Prefix Sum array technique to find all subarray sums in constant time. Follow the steps below to solve the problem:

• Initialize a vector sum of size N to store the prefix sum of the array.
• Initialize two variables, say l & r, to store the two indexes which are to be dropped in order to split the array into 3 equal sum subarrays.
• Sum of the three subarrays would be sum[l – 1], sum[r – 1] – sum[l] and sum[N – 1] – sum[r].
• Iterate over the range [1, N – 4] using a variable l:
• Iterate over the range [l + 2, N – 2] using variable r and check if at any point, left subarray sum is equal to middle subarray sum and right subarray sum, then print the values of l & r and return.
• If no such pair exists, print -1.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to check if array can``// be split into three equal sum``// subarrays by removing a pair``void` `findSplit(``int` `arr[], ``int` `N)``{``    ``// Stores prefix sum array``    ``vector<``int``> sum(N);` `    ``// Copy array elements``    ``for` `(``int` `i = 0; i < N; i++) {``        ``sum[i] = arr[i];``    ``}` `    ``// Traverse the array``    ``for` `(``int` `i = 1; i < N; i++) {``        ``sum[i] += sum[i - 1];``    ``}` `    ``for` `(``int` `l = 1; l <= N - 4; l++) {` `        ``for` `(``int` `r = l + 2; r <= N - 2; r++) {` `            ``// Stores sums of all three subarrays``            ``int` `lsum = 0, rsum = 0, msum = 0;` `            ``// Sum of left subarray``            ``lsum = sum[l - 1];` `            ``// Sum of middle subarray``            ``msum = sum[r - 1] - sum[l];` `            ``// Sum of right subarray``            ``rsum = sum[N - 1] - sum[r];` `            ``// Check if sum of subarrays are equal``            ``if` `(lsum == rsum && rsum == msum) {` `                ``// Print the possible pair``                ``cout << l << ``" "` `<< r << endl;``                ``return``;``            ``}``        ``}``    ``}` `    ``// If no such pair exists, print -1``    ``cout << -1 << endl;``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 2, 5, 12, 7, 19, 4, 3 };` `    ``// Size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``findSplit(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `// Function to check if array can``// be split into three equal sum``// subarrays by removing a pair``static` `void` `findSplit(``int` `arr[], ``int` `N)``{``  ` `    ``// Stores prefix sum array``    ``int` `[]sum = ``new` `int``[N];` `    ``// Copy array elements``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``sum[i] = arr[i];``    ``}` `    ``// Traverse the array``    ``for` `(``int` `i = ``1``; i < N; i++)``    ``{``        ``sum[i] += sum[i - ``1``];``    ``}` `    ``for` `(``int` `l = ``1``; l <= N - ``4``; l++) {` `        ``for` `(``int` `r = l + ``2``; r <= N - ``2``; r++) {` `            ``// Stores sums of all three subarrays``            ``int` `lsum = ``0``, rsum = ``0``, msum = ``0``;` `            ``// Sum of left subarray``            ``lsum = sum[l - ``1``];` `            ``// Sum of middle subarray``            ``msum = sum[r - ``1``] - sum[l];` `            ``// Sum of right subarray``            ``rsum = sum[N - ``1``] - sum[r];` `            ``// Check if sum of subarrays are equal``            ``if` `(lsum == rsum && rsum == msum) {` `                ``// Print the possible pair``                ``System.out.print(l+ ``" "` `+  r +``"\n"``);``                ``return``;``            ``}``        ``}``    ``}` `    ``// If no such pair exists, print -1``    ``System.out.print(-``1` `+``"\n"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ` `    ``// Given array``    ``int` `arr[] = { ``2``, ``5``, ``12``, ``7``, ``19``, ``4``, ``3` `};` `    ``// Size of the array``    ``int` `N = arr.length;``    ``findSplit(arr, N);``}``}` `// This code is contributed by shikhasingrajput`

## Python3

 `# Python3 program for the above approach` `# Function to check if array can``# be split into three equal sum``# subarrays by removing a pair``def` `findSplit(arr, N):``  ` `    ``# Stores prefix sum array``    ``sum` `=` `[i ``for` `i ``in` `arr]` `    ``# Traverse the array``    ``for` `i ``in` `range``(``1``, N):``        ``sum``[i] ``+``=` `sum``[i ``-` `1``]` `    ``for` `l ``in` `range``(``1``, N ``-` `3``):``        ``for` `r ``in` `range``(l ``+` `2``, N ``-` `1``):``          ` `            ``# Stores sums of all three subarrays``            ``lsum , rsum , msum ``=``0``, ``0``, ``0` `            ``# Sum of left subarray``            ``lsum ``=` `sum``[l ``-` `1``]` `            ``# Sum of middle subarray``            ``msum ``=` `sum``[r ``-` `1``] ``-` `sum``[l]` `            ``# Sum of right subarray``            ``rsum ``=` `sum``[N ``-` `1``] ``-` `sum``[r]` `            ``# Check if sum of subarrays are equal``            ``if` `(lsum ``=``=` `rsum ``and` `rsum ``=``=` `msum):` `                ``# Print possible pair``                ``print``(l, r)``                ``return` `    ``# If no such pair exists, print -1``    ``print` `(``-``1``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``# Given array``    ``arr ``=` `[``2``, ``5``, ``12``, ``7``, ``19``, ``4``, ``3` `]` `    ``# Size of the array``    ``N ``=` `len``(arr)` `    ``findSplit(arr, N)` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach``using` `System;``public` `class` `GFG``{` `// Function to check if array can``// be split into three equal sum``// subarrays by removing a pair``static` `void` `findSplit(``int` `[]arr, ``int` `N)``{``  ` `    ``// Stores prefix sum array``    ``int` `[]sum = ``new` `int``[N];` `    ``// Copy array elements``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``sum[i] = arr[i];``    ``}` `    ``// Traverse the array``    ``for` `(``int` `i = 1; i < N; i++)``    ``{``        ``sum[i] += sum[i - 1];``    ``}` `    ``for` `(``int` `l = 1; l <= N - 4; l++) {``        ``for` `(``int` `r = l + 2; r <= N - 2; r++) {` `            ``// Stores sums of all three subarrays``            ``int` `lsum = 0, rsum = 0, msum = 0;` `            ``// Sum of left subarray``            ``lsum = sum[l - 1];` `            ``// Sum of middle subarray``            ``msum = sum[r - 1] - sum[l];` `            ``// Sum of right subarray``            ``rsum = sum[N - 1] - sum[r];` `            ``// Check if sum of subarrays are equal``            ``if` `(lsum == rsum && rsum == msum) {` `                ``// Print the possible pair``                ``Console.Write(l+ ``" "` `+  r +``"\n"``);``                ``return``;``            ``}``        ``}``    ``}` `    ``// If no such pair exists, print -1``    ``Console.Write(-1 +``"\n"``);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ` `    ``// Given array``    ``int` `[]arr = { 2, 5, 12, 7, 19, 4, 3 };` `    ``// Size of the array``    ``int` `N = arr.Length;``    ``findSplit(arr, N);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`2 4`

Time Complexity: O(N2)
Auxiliary Space: O(N)

Most Optimal Approach: The most optimal idea is to make use of the two-pointer technique along with the use of Prefix Sum. Follow the steps below to solve the problem:

• Initialize a vector of size N to store the prefix sum of the array.
• Initialize two variables, say l & r, to traverse the array using the two-pointer approach.
• Traverse the array till l < r or until either all three sums become equal:
• If the sum of the left subarray is greater than the sum of the right subarray, add an extra element to the right subarray. Therefore, reducing the value of r by 1.
• If the sum of the right subarray is greater than the sum of the left subarray, add an element to the left subarray. Therefore, increasing l by 1.
• If both the sum of left and right subarrays are equal, but not equal to the sum of the middle subarray increase l by 1 and reduce r by 1.
• If no such pair exists, print -1.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to check if array can``// be split into three equal sum``// subarrays by removing a pair``void` `findSplit(``int` `arr[], ``int` `N)``{``    ``// Two pointers l and r``    ``int` `l = 1, r = N - 2;``    ``int` `lsum, msum, rsum;` `    ``// Stores prefix sum array``    ``vector<``int``> sum(N);``    ``sum[0] = arr[0];` `    ``// Traverse the array``    ``for` `(``int` `i = 1; i < N; i++) {``        ``sum[i] = sum[i - 1] + arr[i];``    ``}` `    ``// Two pointer approach``    ``while` `(l < r) {` `        ``// Sum of left subarray``        ``lsum = sum[l - 1];` `        ``// Sum of middle subarray``        ``msum = sum[r - 1] - sum[l];` `        ``// Sum of right subarray``        ``rsum = sum[N - 1] - sum[r];` `        ``// Print split indices if sum is equal``        ``if` `(lsum == msum and msum == rsum) {``            ``cout << l << ``" "` `<< r << endl;``            ``return``;``        ``}` `        ``// Move left pointer if lsum < rsum``        ``if` `(lsum < rsum)``            ``l++;` `        ``// Move right pointer if rsum > lsum``        ``else` `if` `(lsum > rsum)``            ``r--;` `        ``// Move both pointers if lsum = rsum``        ``// but they are not equal to msum``        ``else` `{``            ``l++;``            ``r--;``        ``}``    ``}` `    ``// If no possible pair exists, print -1``    ``cout << -1 << endl;``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 2, 5, 12, 7, 19, 4, 3 };` `    ``// Size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``findSplit(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``public` `class` `GFG``{` `  ``// Function to check if array can``  ``// be split into three equal sum``  ``// subarrays by removing a pair``  ``static` `void` `findSplit(``int` `[]arr, ``int` `N)``  ``{` `    ``// Two pointers l and r``    ``int` `l = ``1``, r = N - ``2``;``    ``int` `lsum, msum, rsum;` `    ``// Stores prefix sum array``    ``int` `sum[] = ``new` `int``[N];` `    ``sum[``0``] = arr[``0``];` `    ``// Traverse the array``    ``for` `(``int` `i = ``1``; i < N; i++) {``      ``sum[i] = sum[i - ``1``] + arr[i];``    ``}` `    ``// Two pointer approach``    ``while` `(l < r) {` `      ``// Sum of left subarray``      ``lsum = sum[l - ``1``];` `      ``// Sum of middle subarray``      ``msum = sum[r - ``1``] - sum[l];` `      ``// Sum of right subarray``      ``rsum = sum[N - ``1``] - sum[r];` `      ``// Print split indices if sum is equal``      ``if` `(lsum == msum && msum == rsum) {``        ``System.out.println(l + ``" "` `+ r);``        ``return``;``      ``}` `      ``// Move left pointer if lsum < rsum``      ``if` `(lsum < rsum)``        ``l++;` `      ``// Move right pointer if rsum > lsum``      ``else` `if` `(lsum > rsum)``        ``r--;` `      ``// Move both pointers if lsum = rsum``      ``// but they are not equal to msum``      ``else` `{``        ``l++;``        ``r--;``      ``}``    ``}` `    ``// If no possible pair exists, print -1``    ``System.out.println(-``1``);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main (String[] args)``  ``{``    ``// Given array``    ``int` `[]arr = { ``2``, ``5``, ``12``, ``7``, ``19``, ``4``, ``3` `};` `    ``// Size of the array``    ``int` `N = arr.length;` `    ``findSplit(arr, N);``  ``}``}` `// This code is contributed by AnkThon`

## Python3

 `# Python3 program for the above approach` `# Function to check if array can``# be split into three equal sum``# subarrays by removing a pair``def` `findSplit(arr, N) :` `    ``# Two pointers l and r``    ``l ``=` `1``; r ``=` `N ``-` `2``;` `    ``# Stores prefix sum array``    ``sum` `=` `[``0``]``*``N;``    ``sum``[``0``] ``=` `arr[``0``];` `    ``# Traverse the array``    ``for` `i ``in` `range``(``1``, N) :``        ``sum``[i] ``=` `sum``[i ``-` `1``] ``+` `arr[i];` `    ``# Two pointer approach``    ``while` `(l < r) :` `        ``# Sum of left subarray``        ``lsum ``=` `sum``[l ``-` `1``];` `        ``# Sum of middle subarray``        ``msum ``=` `sum``[r ``-` `1``] ``-` `sum``[l];` `        ``# Sum of right subarray``        ``rsum ``=` `sum``[N ``-` `1``] ``-` `sum``[r];` `        ``# Print split indices if sum is equal``        ``if` `(lsum ``=``=` `msum ``and` `msum ``=``=` `rsum) :``            ``print``(l,r);``            ``return``;` `        ``# Move left pointer if lsum < rsum``        ``if` `(lsum < rsum) :``            ``l ``+``=` `1``;` `        ``# Move right pointer if rsum > lsum``        ``elif` `(lsum > rsum) :``            ``r ``-``=` `1``;` `        ``# Move both pointers if lsum = rsum``        ``# but they are not equal to msum``        ``else` `:``            ``l ``+``=` `1``;``            ``r ``-``=` `1``;` `    ``# If no possible pair exists, print -1``    ``print``(``-``1``);` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``# Given array``    ``arr ``=` `[ ``2``, ``5``, ``12``, ``7``, ``19``, ``4``, ``3` `];` `    ``# Size of the array``    ``N ``=` `len``(arr);` `    ``findSplit(arr, N);` `    ``# This code is contributed by AnkThon`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG {``    ` `    ``// Function to check if array can``    ``// be split into three equal sum``    ``// subarrays by removing a pair``    ``static` `void` `findSplit(``int``[] arr, ``int` `N)``    ``{`` ` `      ``// Two pointers l and r``      ``int` `l = 1, r = N - 2;``      ``int` `lsum, msum, rsum;`` ` `      ``// Stores prefix sum array``      ``int``[] sum = ``new` `int``[N];`` ` `      ``sum[0] = arr[0];`` ` `      ``// Traverse the array``      ``for` `(``int` `i = 1; i < N; i++) {``        ``sum[i] = sum[i - 1] + arr[i];``      ``}`` ` `      ``// Two pointer approach``      ``while` `(l < r) {`` ` `        ``// Sum of left subarray``        ``lsum = sum[l - 1];`` ` `        ``// Sum of middle subarray``        ``msum = sum[r - 1] - sum[l];`` ` `        ``// Sum of right subarray``        ``rsum = sum[N - 1] - sum[r];`` ` `        ``// Print split indices if sum is equal``        ``if` `(lsum == msum && msum == rsum) {``          ``Console.Write(l + ``" "` `+ r);``          ``return``;``        ``}`` ` `        ``// Move left pointer if lsum < rsum``        ``if` `(lsum < rsum)``          ``l++;`` ` `        ``// Move right pointer if rsum > lsum``        ``else` `if` `(lsum > rsum)``          ``r--;`` ` `        ``// Move both pointers if lsum = rsum``        ``// but they are not equal to msum``        ``else` `{``          ``l++;``          ``r--;``        ``}``      ``}`` ` `      ``// If no possible pair exists, print -1``      ``Console.Write(-1);``    ``}``    ` `  ``static` `void` `Main()``  ``{``    ` `    ``// Given array``    ``int``[] arr = { 2, 5, 12, 7, 19, 4, 3 };``  ` `    ``// Size of the array``    ``int` `N = arr.Length;``  ` `    ``findSplit(arr, N);``  ``}``}` `// This code is contributed by rameshtravel07.`

## Javascript

 ``
Output:
`2 4`

Time Complexity: O(N)
Auxiliary Space: O(N)

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