Split an array into two equal Sum subarrays
Last Updated : 11 Jul, 2022
Given an array of integers greater than zero, find if it is possible to split it in two subarrays (without reordering the elements), such that the sum of the two subarrays is the same. Print the two subarrays.
Examples :
Input : Arr[] = { 1 , 2 , 3 , 4 , 5 , 5 } Output : { 1 2 3 4 } { 5 , 5 } Input : Arr[] = { 4, 1, 2, 3 } Output : {4 1} {2 3} Input : Arr[] = { 4, 3, 2, 1} Output : Not Possible Asked In : Facebook interview
A Simple solution is to run two loop to split array and check it is possible to split array into two parts such that sum of first_part equal to sum of second_part.
Below is the implementation of above idea.
C++ // C++ program to split an array into Two // equal sum subarrays #include<bits/stdc++.h> using namespace std; // Returns split point. If not possible, then // return -1. int findSplitPoint(int arr[], int n) { int leftSum = 0 ; // traverse array element for (int i = 0; i < n; i++) { // add current element to left Sum leftSum += arr[i] ; // find sum of rest array elements (rightSum) int rightSum = 0 ; for (int j = i+1 ; j < n ; j++ ) rightSum += arr[j] ; // split point index if (leftSum == rightSum) return i+1 ; } // if it is not possible to split array into // two parts return -1; } // Prints two parts after finding split point using // findSplitPoint() void printTwoParts(int arr[], int n) { int splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { cout << "Not Possible" <<endl; return; } for (int i = 0; i < n; i++) { if(splitPoint == i) cout << endl; cout << arr[i] << " " ; } } // driver program int main() { int arr[] = {1 , 2 , 3 , 4 , 5 , 5 }; int n = sizeof(arr)/sizeof(arr[0]); printTwoParts(arr, n); return 0; } // Java program to split an array // into two equal sum subarrays import java.io.*; class GFG { // Returns split point. If // not possible, then return -1. static int findSplitPoint(int arr[], int n) { int leftSum = 0 ; // traverse array element for (int i = 0; i < n; i++) { // add current element to left Sum leftSum += arr[i] ; // find sum of rest array // elements (rightSum) int rightSum = 0 ; for (int j = i+1 ; j < n ; j++ ) rightSum += arr[j] ; // split point index if (leftSum == rightSum) return i+1 ; } // if it is not possible to // split array into two parts return -1; } // Prints two parts after finding // split point using findSplitPoint() static void printTwoParts(int arr[], int n) { int splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { System.out.println("Not Possible"); return; } for (int i = 0; i < n; i++) { if(splitPoint == i) System.out.println(); System.out.print(arr[i] + " "); } } // Driver program public static void main (String[] args) { int arr[] = {1 , 2 , 3 , 4 , 5 , 5 }; int n = arr.length; printTwoParts(arr, n); } } // This code is contributed by vt_m # Python3 program to split an array into Two # equal sum subarrays # Returns split point. If not possible, then # return -1. def findSplitPoint(arr, n) : leftSum = 0 # traverse array element for i in range(0, n) : # add current element to left Sum leftSum += arr[i] # find sum of rest array elements (rightSum) rightSum = 0 for j in range(i+1, n) : rightSum += arr[j] # split point index if (leftSum == rightSum) : return i+1 # if it is not possible to split array into # two parts return -1 # Prints two parts after finding split point using # findSplitPoint() def printTwoParts(arr, n) : splitPo = findSplitPoint(arr, n) if (splitPo == -1 or splitPo == n ) : print ("Not Possible") return for i in range(0, n) : if(splitPo == i) : print ("") print (str(arr[i]) + ' ',end='') # driver program arr = [1 , 2 , 3 , 4 , 5 , 5] n = len(arr) printTwoParts(arr, n) # This code is contributed by Manish Shaw # (manishshaw1) // C# program to split an array // into two equal sum subarrays using System; class GFG { // Returns split point. If // not possible, then return -1. static int findSplitPoint(int []arr, int n) { int leftSum = 0 ; // traverse array element for (int i = 0; i < n; i++) { // add current element to left Sum leftSum += arr[i] ; // find sum of rest array // elements (rightSum) int rightSum = 0 ; for (int j = i+1 ; j < n ; j++ ) rightSum += arr[j] ; // split point index if (leftSum == rightSum) return i+1 ; } // if it is not possible to // split array into two parts return -1; } // Prints two parts after finding // split point using findSplitPoint() static void printTwoParts(int []arr, int n) { int splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { Console.Write("Not Possible"); return; } for (int i = 0; i < n; i++) { if(splitPoint == i) Console.WriteLine(); Console.Write(arr[i] + " "); } } // Driver program public static void Main () { int []arr = {1 , 2 , 3 , 4 , 5 , 5 }; int n = arr.Length; printTwoParts(arr, n); } } // This code is contributed by nitin mittal <?php // PHP program to split // an array into Two // equal sum subarrays // Returns split point. // If not possible, then // return -1. function findSplitPoint( $arr, $n) { $leftSum = 0 ; // traverse array element for($i = 0; $i < $n; $i++) { // add current element // to left Sum $leftSum += $arr[$i] ; // find sum of rest array // elements (rightSum) $rightSum = 0 ; for($j = $i + 1 ; $j < $n ; $j++ ) $rightSum += $arr[$j] ; // split point index if ($leftSum == $rightSum) return $i+1 ; } // if it is not possible // to split array into // two parts return -1; } // Prints two parts after // finding split point using // findSplitPoint() function printTwoParts($arr, $n) { $splitPoint = findSplitPoint($arr, $n); if ($splitPoint == -1 or $splitPoint == $n ) { echo "Not Possible" ; return; } for ( $i = 0; $i < $n; $i++) { if($splitPoint == $i) echo "\n"; echo $arr[$i] , " " ; } } // Driver Code $arr = array(1 , 2 , 3 , 4 , 5 , 5); $n = count($arr); printTwoParts($arr, $n); // This code is contributed by anuj_67. ?> <script> // Java script program to split an array // into two equal sum subarrays // Returns split point. If // not possible, then return -1. function findSplitPoint(arr,n) { let leftSum = 0 ; // traverse array element for (let i = 0; i < n; i++) { // add current element to left Sum leftSum += arr[i] ; // find sum of rest array // elements (rightSum) let rightSum = 0 ; for (let j = i+1 ; j < n ; j++ ) rightSum += arr[j] ; // split point index if (leftSum == rightSum) return i+1 ; } // if it is not possible to // split array into two parts return -1; } // Prints two parts after finding // split point using findSplitPoint() function printTwoParts(arr,n) { let splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { document.write("Not Possible"); return; } for (let i = 0; i < n; i++) { if(splitPoint == i) document.write("<br>"); document.write(arr[i] + " "); } } // Driver program let arr = [1 , 2 , 3 , 4 , 5 , 5 ]; let n = arr.length; printTwoParts(arr, n); // contributed by sravan kumar </script> Time Complexity : O(n2)
Auxiliary Space : O(1)
An Efficient solution is to first compute the sum of the whole array from left to right. Now we traverse array from right and keep track of right sum, left sum can be computed by subtracting current element from whole sum.
Below is the implementation of above idea.
C++ // C++ program to split an array into Two // equal sum subarrays #include<bits/stdc++.h> using namespace std; // Returns split point. If not possible, then // return -1. int findSplitPoint(int arr[], int n) { // traverse array element and compute sum // of whole array int leftSum = 0; for (int i = 0 ; i < n ; i++) leftSum += arr[i]; // again traverse array and compute right sum // and also check left_sum equal to right // sum or not int rightSum = 0; for (int i=n-1; i >= 0; i--) { // add current element to right_sum rightSum += arr[i]; // exclude current element to the left_sum leftSum -= arr[i] ; if (rightSum == leftSum) return i ; } // if it is not possible to split array // into two parts. return -1; } // Prints two parts after finding split point using // findSplitPoint() void printTwoParts(int arr[], int n) { int splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { cout << "Not Possible" <<endl; return; } for (int i = 0; i < n; i++) { if(splitPoint == i) cout << endl; cout << arr[i] << " " ; } } // driver program int main() { int arr[] = {1 , 2 , 3 , 4 , 5 , 5 }; int n = sizeof(arr)/sizeof(arr[0]); printTwoParts(arr, n); return 0; } // java program to split an array // into Two equal sum subarrays import java.io.*; class GFG { // Returns split point. If not possible, then // return -1. static int findSplitPoint(int arr[], int n) { // traverse array element and compute sum // of whole array int leftSum = 0; for (int i = 0 ; i < n ; i++) leftSum += arr[i]; // again traverse array and compute right // sum and also check left_sum equal to // right sum or not int rightSum = 0; for (int i = n-1; i >= 0; i--) { // add current element to right_sum rightSum += arr[i]; // exclude current element to the left_sum leftSum -= arr[i] ; if (rightSum == leftSum) return i ; } // if it is not possible to split array // into two parts. return -1; } // Prints two parts after finding split // point using findSplitPoint() static void printTwoParts(int arr[], int n) { int splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { System.out.println("Not Possible" ); return; } for (int i = 0; i < n; i++) { if(splitPoint == i) System.out.println(); System.out.print(arr[i] + " "); } } // Driver program public static void main (String[] args) { int arr[] = {1 , 2 , 3 , 4 , 5 , 5 }; int n = arr.length; printTwoParts(arr, n); } } // This code is contributed by vt_m # Python3 program to split # an array into Two # equal sum subarrays # Returns split point. # If not possible, # then return -1. def findSplitPoint(arr, n) : # traverse array element and # compute sum of whole array leftSum = 0 for i in range(0, n) : leftSum += arr[i] # again traverse array and # compute right sum and also # check left_sum equal to # right sum or not rightSum = 0 for i in range(n-1, -1, -1) : # add current element # to right_sum rightSum += arr[i] # exclude current element # to the left_sum leftSum -= arr[i] if (rightSum == leftSum) : return i # if it is not possible # to split array into # two parts. return -1 # Prints two parts after # finding split point # using findSplitPoint() def printTwoParts(arr, n) : splitPoint = findSplitPoint(arr, n) if (splitPoint == -1 or splitPoint == n ) : print ("Not Possible") return for i in range (0, n) : if(splitPoint == i) : print ("") print (arr[i], end = " ") # Driver Code arr = [1, 2, 3, 4, 5, 5] n = len(arr) printTwoParts(arr, n) # This code is contributed by Manish Shaw # (manishshaw1) // C# program to split an array // into Two equal sum subarrays using System; class GFG { // Returns split point. If not possible, then // return -1. static int findSplitPoint(int []arr, int n) { // traverse array element and compute sum // of whole array int leftSum = 0; for (int i = 0 ; i < n ; i++) leftSum += arr[i]; // again traverse array and compute right // sum and also check left_sum equal to // right sum or not int rightSum = 0; for (int i = n-1; i >= 0; i--) { // add current element to right_sum rightSum += arr[i]; // exclude current element to the left_sum leftSum -= arr[i] ; if (rightSum == leftSum) return i ; } // if it is not possible to split array // into two parts. return -1; } // Prints two parts after finding split // point using findSplitPoint() static void printTwoParts(int []arr, int n) { int splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { Console.Write("Not Possible" ); return; } for (int i = 0; i < n; i++) { if(splitPoint == i) Console.WriteLine(); Console.Write(arr[i] + " "); } } // Driver program public static void Main (String[] args) { int []arr = {1 , 2 , 3 , 4 , 5 , 5 }; int n = arr.Length; printTwoParts(arr, n); } } // This code is contributed by parashar <?php // PHP program to split // an array into Two // equal sum subarrays // Returns split point. // If not possible, // then return -1. function findSplitPoint($arr, $n) { // traverse array element and // compute sum of whole array $leftSum = 0; for ( $i = 0 ; $i < $n ; $i++) $leftSum += $arr[$i]; // again traverse array and // compute right sum and also // check left_sum equal to // right sum or not $rightSum = 0; for ($i = $n - 1; $i >= 0; $i--) { // add current element // to right_sum $rightSum += $arr[$i]; // exclude current element // to the left_sum $leftSum -= $arr[$i] ; if ($rightSum == $leftSum) return $i ; } // if it is not possible // to split array into // two parts. return -1; } // Prints two parts after // finding split point // using findSplitPoint() function printTwoParts( $arr, $n) { $splitPoint = findSplitPoint($arr, $n); if ($splitPoint == -1 or $splitPoint == $n ) { echo "Not Possible" ; return; } for ( $i = 0; $i < $n; $i++) { if($splitPoint == $i) echo "\n"; echo $arr[$i] , " " ; } } // Driver Code $arr = array(1, 2, 3, 4, 5, 5); $n = count($arr); printTwoParts($arr, $n); // This code is contributed by anuj_67. ?> <script> // Javascript program to split an array // into Two equal sum subarrays // Returns split point. If not possible, then // return -1. function findSplitPoint(arr, n) { // traverse array element and compute sum // of whole array let leftSum = 0; for (let i = 0 ; i < n ; i++) leftSum += arr[i]; // again traverse array and compute right // sum and also check left_sum equal to // right sum or not let rightSum = 0; for (let i = n-1; i >= 0; i--) { // add current element to right_sum rightSum += arr[i]; // exclude current element to the left_sum leftSum -= arr[i] ; if (rightSum == leftSum) return i ; } // if it is not possible to split array // into two parts. return -1; } // Prints two parts after finding split // point using findSplitPoint() function printTwoParts(arr, n) { let splitPoint = findSplitPoint(arr, n); if (splitPoint == -1 || splitPoint == n ) { document.write("Not Possible" ); return; } for (let i = 0; i < n; i++) { if(splitPoint == i) document.write("</br>"); document.write(arr[i] + " "); } } let arr = [1 , 2 , 3 , 4 , 5 , 5 ]; let n = arr.length; printTwoParts(arr, n); // This code is contributed by rameshtravel07. </script> Time Complexity : O(n)
Auxiliary Space : O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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