Find an element in array such that sum of left array is equal to sum of right array

Given, an array of size n. Find an element which divides the array in two sub-arrays with equal sum.

Examples:

Input : 1 4 2 5
Output : 2
Explanation : If 2 is the partition, 
      subarrays are : {1, 4} and {5}

Input : 2 3 4 1 4 5
Output : 1
Explanation : If 1 is the partition, 
 Subarrays are : {2, 3, 4} and {4, 5}

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Simple)
Consider every element starting from second element. Compute sum of elements on its left and sum of elements on its right. If these two sums are same, return the element.

Method 2 (Using Prefix and Suffx Arrays :

We form a prefix and suffix sum arrays

Given array : 1 4 2 5
Prefix Sum :  1  5 7 12
Suffix Sum :  12 11 7 5

Now, we will traverse both prefix arrays.
The index at which they yield equal result,
is the index where the array is partitioned 
with equal sum.

Implementation :

C++

#include <bits/stdc++.h> 
using namespace std; 
  
int findElement(int arr[], int n) 
{ 
    // Forming prefix sum array from 0 
    int prefixSum[n]; 
    prefixSum[0] = arr[0]; 
    for (int i = 1; i < n; i++) 
        prefixSum[i] = prefixSum[i - 1] + arr[i]; 
  
    // Forming suffix sum array from n-1 
    int suffixSum[n]; 
    suffixSum[n - 1] = arr[n - 1]; 
    for (int i = n - 2; i >= 0; i--) 
        suffixSum[i] = suffixSum[i + 1] + arr[i]; 
  
    // Find the point where prefix and suffix 
    // sums are same. 
    for (int i = 1; i < n - 1; i++) 
        if (prefixSum[i] == suffixSum[i]) 
            return arr[i]; 
  
    return -1; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 1, 4, 2, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << findElement(arr, n); 
    return 0; 
} 

Java

// Java program to find an element  
// such that sum of right side element  
// is equal to sum of left side 
public class GFG { 
      
    // Finds an element in an array such that 
    // left and right side sums are equal 
    static int findElement(int arr[], int n) 
    { 
        // Forming prefix sum array from 0 
        int[] prefixSum = new int[n]; 
        prefixSum[0] = arr[0]; 
        for (int i = 1; i < n; i++) 
            prefixSum[i] = prefixSum[i - 1] + arr[i]; 
       
        // Forming suffix sum array from n-1 
        int[] suffixSum = new int[n]; 
        suffixSum[n - 1] = arr[n - 1]; 
        for (int i = n - 2; i >= 0; i--) 
            suffixSum[i] = suffixSum[i + 1] + arr[i]; 
       
        // Find the point where prefix and suffix 
        // sums are same. 
        for (int i = 1; i < n - 1; i++) 
            if (prefixSum[i] == suffixSum[i]) 
                return arr[i]; 
       
        return -1; 
    } 
       
    // Driver code 
    public static void main(String args[]) 
    { 
        int arr[] = { 1, 4, 2, 5 }; 
        int n = arr.length; 
        System.out.println(findElement(arr, n)); 
    } 
} 
// This code is contributed by Sumit Ghosh 

Python 3

# Python 3 Program to find an element  
# such that sum of right side element  
# is equal to sum of left side 
  
# Function for Finds an element in  
# an array such that left and right 
# side sums are equal  
def findElement(arr, n) : 
      
    # Forming prefix sum array from 0  
    prefixSum = [0] * n 
    prefixSum[0] = arr[0] 
    for i in range(1, n) : 
        prefixSum[i] = prefixSum[i - 1] + arr[i] 
  
    # Forming suffix sum array from n-1 
    suffixSum = [0] * n 
    suffixSum[n - 1] = arr[n - 1] 
    for i in range(n - 2, -1, -1) : 
        suffixSum[i] = suffixSum[i + 1] + arr[i] 
  
    # Find the point where prefix  
    # and suffix sums are same. 
    for i in range(1, n - 1, 1) : 
        if prefixSum[i] == suffixSum[i] : 
            return arr[i] 
          
    return -1
  
# Driver Code 
if __name__ == "__main__" : 
      
    arr = [ 1, 4, 2, 5] 
    n = len(arr) 
    print(findElement(arr, n)) 
  
# This code is contributed by ANKITRAI1 

C#

// C# program to find an element  
// such that sum of right side element  
// is equal to sum of left side 
using System; 
  
class GFG  
{ 
      
    // Finds an element in an  
    // array such that left  
    // and right side sums  
    // are equal 
    static int findElement(int []arr,  
                           int n) 
    { 
        // Forming prefix sum 
        // array from 0 
        int[] prefixSum = new int[n]; 
        prefixSum[0] = arr[0]; 
        for (int i = 1; i < n; i++) 
            prefixSum[i] = prefixSum[i - 1] +  
                                     arr[i]; 
      
        // Forming suffix sum  
        // array from n-1 
        int[] suffixSum = new int[n]; 
        suffixSum[n - 1] = arr[n - 1]; 
        for (int i = n - 2; i >= 0; i--) 
            suffixSum[i] = suffixSum[i + 1] +  
                                     arr[i]; 
      
        // Find the point where prefix  
        // and suffix sums are same. 
        for (int i = 1; i < n - 1; i++) 
            if (prefixSum[i] == suffixSum[i]) 
                return arr[i]; 
      
        return -1; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int []arr = { 1, 4, 2, 5 }; 
        int n = arr.Length; 
        Console.WriteLine(findElement(arr, n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php  
function findElement(&$arr, $n) 
{ 
    // Forming prefix sum array from 0 
    $prefixSum = array_fill(0, $n, NULL); 
    $prefixSum[0] = $arr[0]; 
    for ($i = 1; $i < $n; $i++) 
        $prefixSum[$i] = $prefixSum[$i - 1] + 
                                    $arr[$i]; 
  
    // Forming suffix sum array from n-1 
    $suffixSum = array_fill(0, $n, NULL); 
    $suffixSum[$n - 1] = $arr[$n - 1]; 
    for ($i = $n - 2; $i >= 0; $i--) 
        $suffixSum[$i] = $suffixSum[$i + 1] +  
                                    $arr[$i]; 
  
    // Find the point where prefix  
    // and suffix sums are same. 
    for ($i = 1; $i < $n - 1; $i++) 
        if ($prefixSum[$i] == $suffixSum[$i]) 
            return $arr[$i]; 
  
    return -1; 
} 
  
// Driver code 
$arr = array( 1, 4, 2, 5 ); 
$n = sizeof($arr); 
echo findElement($arr, $n); 
  
// This code is contributed  
// by ChitraNayal 
?> 

Output:

Method 3 (Space efficient)
We calculate the sum of the whole array except the first element in right_sum, considering it to be the partitioning element. Now, we traverse the array from left to right, subtracting an element from right_sum and adding an element to left_sum. The point where right_sum equals left_sum, we get the partition.

Below is the implementation :

C++

#include <bits/stdc++.h> 
using namespace std; 
  
// Function to compute partition 
int findElement(int arr[], int size) 
{ 
    int right_sum = 0, left_sum = 0; 
  
    // Computing right_sum 
    for (int i = 1; i < size; i++) 
        right_sum += arr[i]; 
  
    // Checking the point of partition 
    // i.e. left_Sum == right_sum 
    for (int i = 0, j = 1; j < size; i++, j++) { 
        right_sum -= arr[j]; 
        left_sum += arr[i]; 
  
        if (left_sum == right_sum) 
            return arr[i + 1]; 
    } 
  
    return -1; 
} 
  
// Driver 
int main() 
{ 
    int arr[] = { 2, 3, 4, 1, 4, 5 }; 
    int size = sizeof(arr) / sizeof(arr[0]); 
    cout << findElement(arr, size); 
    return 0; 
} 

Java

// Java program to find an element  
// such that sum of right side element  
// is equal to sum of left side 
public class GFG { 
      
    // Function to compute partition 
    static int findElement(int arr[], int size) 
    { 
        int right_sum = 0, left_sum = 0; 
       
        // Computing right_sum 
        for (int i = 1; i < size; i++) 
            right_sum += arr[i]; 
       
        // Checking the point of partition 
        // i.e. left_Sum == right_sum 
        for (int i = 0, j = 1; j < size; i++, j++) { 
            right_sum -= arr[j]; 
            left_sum += arr[i]; 
       
            if (left_sum == right_sum) 
                return arr[i + 1]; 
        } 
       
        return -1; 
    } 
       
    // Driver 
    public static void main(String args[]) 
    { 
        int arr[] = { 2, 3, 4, 1, 4, 5 }; 
        int size = arr.length; 
        System.out.println(findElement(arr, size)); 
    } 
} 
// This code is contributed by Sumit Ghosh 

Python 3

# Python 3 Program to find an element  
# such that sum of right side element  
# is equal to sum of left side 
  
# Function to compute partition 
def findElement(arr, size) : 
  
    right_sum, left_sum = 0, 0
  
    # Computing right_sum 
    for i in range(1, size) : 
        right_sum += arr[i] 
  
    i, j = 0, 1
  
    # Checking the point of partition  
    # i.e. left_Sum == right_sum  
    while j < size : 
        right_sum -= arr[j] 
        left_sum += arr[i] 
  
        if left_sum == right_sum : 
            return arr[i + 1] 
  
        j += 1
        i += 1
  
    return -1
  
# Driver Code 
if __name__ == "__main__" : 
      
    arr = [ 2, 3, 4, 1, 4, 5] 
    n = len(arr) 
    print(findElement(arr, n)) 
  
# This code is contributed by ANKITRAI1 

C#

// C# program to find an  
// element such that sum  
// of right side element  
// is equal to sum of  
// left side 
using System; 
  
class GFG  
{ 
    // Function to compute  
    // partition 
    static int findElement(int []arr,  
                           int size) 
    { 
        int right_sum = 0,  
            left_sum = 0; 
      
        // Computing right_sum 
        for (int i = 1; i < size; i++) 
            right_sum += arr[i]; 
      
        // Checking the point  
        // of partition i.e.  
        // left_Sum == right_sum 
        for (int i = 0, j = 1;  
                 j < size; i++, j++) 
        { 
            right_sum -= arr[j]; 
            left_sum += arr[i]; 
      
            if (left_sum == right_sum) 
                return arr[i + 1]; 
        } 
      
        return -1; 
    } 
      
    // Driver Code 
    public static void Main() 
    { 
        int []arr = {2, 3, 4, 1, 4, 5}; 
        int size = arr.Length; 
        Console.WriteLine(findElement(arr, size)); 
    } 
} 
  
// This code is contributed  
// by anuj_67. 

PHP

<?php  
// Function to compute partition 
function findElement(&$arr, $size) 
{ 
    $right_sum = 0; 
    $left_sum = 0; 
  
    // Computing right_sum 
    for ($i = 1; $i < $size; $i++) 
        $right_sum += $arr[$i]; 
  
    // Checking the point of partition 
    // i.e. left_Sum == right_sum 
    for ($i = 0, $j = 1;  
         $j < $size; $i++, $j++)  
    { 
        $right_sum -= $arr[$j]; 
        $left_sum += $arr[$i]; 
  
        if ($left_sum == $right_sum) 
            return $arr[$i + 1]; 
    } 
  
    return -1; 
} 
  
// Driver Code 
$arr = array( 2, 3, 4, 1, 4, 5 ); 
$size = sizeof($arr); 
echo findElement($arr, $size); 
  
// This code is contributed  
// by ChitraNayal 
?> 

Output:

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

C++

Java

Python 3

C#

PHP

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