# Split an array into two equal Sum subarrays

• Difficulty Level : Easy
• 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

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#includeusing 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" <

## Java

 // Java program to split an array // into two equal sum subarraysimport 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

 # 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 programarr = [1 , 2 , 3 , 4 , 5 , 5]n = len(arr)printTwoParts(arr, n)  # This code is contributed by Manish Shaw# (manishshaw1)

## C#

 // C# program to split an array // into two equal sum subarraysusing 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



## Javascript



Output

1 2 3 4
5 5

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#includeusing 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" <

## Java

 // java program to split an array // into Two equal sum subarraysimport 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

 # 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 Codearr = [1, 2, 3, 4, 5, 5]n = len(arr)printTwoParts(arr, n)   # This code is contributed by Manish Shaw# (manishshaw1)

## C#

 // C# program to split an array // into Two equal sum subarraysusing 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

 = 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.?>

## Javascript



Output

1 2 3 4
5 5

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

Related Topic: Subarrays, Subsequences, and Subsets in Array

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