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

Difficulty Level : Easy
Last Updated : 05 May, 2021

Given, an array of size n. Find an element that divides the array into two sub-arrays with equal sums.
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}

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

Method 2 (Using Prefix and Suffix 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.

Below is the implementation of the above approach:

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
?>

Javascript

<script>
 
// Javascript program to find an element
// such that sum of right side element
// is equal to sum of left side
 
    // Finds an element in an array such that
    // left and right side sums are equal
    function findElement(arr , n)
    {
        // Forming prefix sum array from 0
        var prefixSum = Array(n).fill(0);
        prefixSum[0] = arr[0];
        for (i = 1; i < n; i++)
            prefixSum[i] = prefixSum[i - 1] + arr[i];
 
        // Forming suffix sum array from n-1
        var suffixSum = Array(n).fill(0);
        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
     
        var arr = [ 1, 4, 2, 5 ];
        var n = arr.length;
        document.write(findElement(arr, n));
 
// This code contributed by umadevi9616
 
</script>

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. At 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
?>

Javascript

<script>
     // Function to compute partition
    function findElement(arr) {
        let right_sum = 0, left_sum = 0;
  
        // Computing right_sum
        for (let i = 1; i < arr.length; i++)
            right_sum += arr[i];
  
        // Checking the point of partition
        // i.e. left_Sum == right_sum
        for (let i = 0, j = 1; j < arr.length; i++, j++) {
            right_sum -= arr[j];
            left_sum += arr[i];
  
            if (left_sum === right_sum)
                return arr[i + 1];
        }
  
        return -1;
    }
  
    // Driver
    let arr = [ 2, 3, 4, 1, 4, 5 ];
    document.write(findElement(arr));
    
 
// This code is contributed by Surbhi Tyagi
</script>

Output

Method 4 (Both Time and Space Efficient)

We define two pointers i and j to traverse the array from left and right,
left_sum and right_sum to store sum from right and left respectively

If left_sum is lesser than increment i and if right_sum is lesser than decrement j 
and, find a position where left_sum == right_sum and i and j are next to each other

Note: This solution is only applicable if the array contains only positive elements.

Below is the implementation of the above approach:

C++

// C++ program to find an element
// such that sum of right side element
// is equal to sum of left side
#include<bits/stdc++.h>
using namespace std;
     
// Function to compute
// partition
int findElement(int arr[],
                int size)
{
  int right_sum = 0,
      left_sum = 0;
   
  // Maintains left
  // cumulative sum
  left_sum = 0;
 
  // Maintains right
  // cumulative sum
  right_sum = 0;
  int i = -1, j = -1;
 
  for(i = 0, j = size - 1;
      i < j; i++, j--)
  {
    left_sum += arr[i];
    right_sum += arr[j];
 
    // Keep moving i towards
    // center until left_sum
    //is found lesser than right_sum
    while(left_sum < right_sum &&
          i < j)
    {
      i++;
      left_sum += arr[i];
    }
     
    // Keep moving j towards
    // center until right_sum is
    // found lesser than left_sum
    while(right_sum < left_sum &&
          i < j)
    {
      j--;
      right_sum += arr[j];
    }
  }
   
  if(left_sum == right_sum && i == j)
    return arr[i];
  else
    return -1;
}
 
// Driver code
int main()
{
  int arr[] = {2, 3, 4,
               1, 4, 5};
  int size = sizeof(arr) /
             sizeof(arr[0]);
  cout << (findElement(arr, size));
}
 
// This code is contributed by shikhasingrajput

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;
        // Maintains left cumulative sum
        left_sum = 0;
         
        // Maintains right cumulative sum
        right_sum=0;
        int i = -1, j = -1;
         
        for( i = 0, j = size-1 ; i < j ; i++, j-- ){
            left_sum += arr[i];
            right_sum += arr[j];
             
            // Keep moving i towards center until
            // left_sum is found lesser than right_sum
            while(left_sum < right_sum && i < j){
                i++;
                left_sum += arr[i];
            }
            // Keep moving j towards center until
            // right_sum is found lesser than left_sum
            while(right_sum < left_sum && i < j){
                j--;
                right_sum += arr[j];
            }
        }
        if(left_sum == right_sum && i == j)
            return arr[i];
        else
            return -1;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = {2, 3, 4, 1, 4, 5};
        int size = arr.length;
        System.out.println(findElement(arr, size));
    }
}

Python3

# Python3 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):
   
    # Maintains left cumulative sum
    left_sum = 0;
 
    # Maintains right cumulative sum
    right_sum = 0;
    i = 0; j = -1;
    j = size - 1;
     
    while(i < j):
        if(i < j):
            left_sum += arr[i];
            right_sum += arr[j];
 
            # Keep moving i towards center
            # until left_sum is found
            # lesser than right_sum
            while (left_sum < right_sum and
                   i < j):
                i += 1;
                left_sum += arr[i];
 
            # Keep moving j towards center
            # until right_sum is found
            # lesser than left_sum
            while (right_sum < left_sum and
                   i < j):
                j -= 1;
                right_sum += arr[j];
            j -= 1
            i += 1
    if (left_sum == right_sum && i == j):
        return arr[i];
    else:
        return -1;
 
# Driver code
if __name__ == '__main__':
   
    arr = [2, 3, 4,
           1, 4, 5];
    size = len(arr);
    print(findElement(arr, size));
 
# This code is contributed by shikhasingrajput

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;
     
    // Maintains left cumulative sum
    left_sum = 0;
     
    // Maintains right cumulative sum
    right_sum = 0;
    int i = -1, j = -1;
     
    for(i = 0, j = size - 1; i < j; i++, j--)
    {
        left_sum += arr[i];
        right_sum += arr[j];
         
        // Keep moving i towards center until
        // left_sum is found lesser than right_sum
        while (left_sum < right_sum && i < j)
        {
            i++;
            left_sum += arr[i];
        }
         
        // Keep moving j towards center until
        // right_sum is found lesser than left_sum
        while (right_sum < left_sum && i < j)
        {
            j--;
            right_sum += arr[j];
        }
    }
    if (left_sum == right_sum && i == j)
        return arr[i];
    else
        return -1;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 2, 3, 4, 1, 4, 5 };
    int size = arr.Length;
     
    Console.WriteLine(findElement(arr, size));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
// javascript program to find an element
// such that sum of right side element
// is equal to sum of left side
 
    // Function to compute partition
    function findElement(arr , size) {
        var right_sum = 0, left_sum = 0;
        // Maintains left cumulative sum
        left_sum = 0;
 
        // Maintains right cumulative sum
        right_sum = 0;
        var i = -1, j = -1;
 
        for (i = 0, j = size - 1; i < j; i++, j--) {
            left_sum += arr[i];
            right_sum += arr[j];
 
            // Keep moving i towards center until
            // left_sum is found lesser than right_sum
            while (left_sum < right_sum && i < j) {
                i++;
                left_sum += arr[i];
            }
            // Keep moving j towards center until
            // right_sum is found lesser than left_sum
            while (right_sum < left_sum && i < j) {
                j--;
                right_sum += arr[j];
            }
        }
        if (left_sum == right_sum && i == j)
            return arr[i];
        else
            return -1;
    }
 
    // Driver code
     
        var arr = [ 2, 3, 4, 1, 4, 5 ];
        var size = arr.length;
        document.write(findElement(arr, size));
 
// This code contributed by gauravrajput1
</script>

Output

Since all loops start traversing from the last updated i and j pointers and do not cross each other, they run n times in the end.

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

Method 5 (The efficient algorithm):

Here we define two pointers to the array -> start = 0, end = n-1
Two variables to take care of sum -> left_sum = 0, right_sum = 0

Here our algorithm goes like this:

We initialize for loop till the entire size of the array
Basically we check if left_sum > right_sum => add the current end element to the right_sum and decrement end
If right_sum < left_sum => add the current start element to the left_sum and increment start
By these two conditions, we make sure that left_sum and right_sum are nearly balanced, so we can arrive at our solution easily
To make this work during all test cases we need to add a couple of conditional statements to our logic:
If the equilibrium element is found our start will be equal to the end variable and left_sum will be equal right_sum => return the equilibrium element (here we say start == end because we increment/ decrement the pointer after adding the current start/ end value to respective sum. So if, equilibrium element is found start and end should be at the same location)
If start is equal to end variable but left_sum is not equal right_sum => no equilibrium element return -1
If left_sum is equal right_sum, but start is not equal to end => we are still in the middle of the algorithm even though we found that left_sum is equal right_sum we haven’t got that one required equilibrium element (So, in this case, add the current end element to the right_sum and decrement end (or) add the current start element to the left_sum and increment start, to make our algorithm continue further).
Even here there is one test case that needs to be handled:
When there is only one element in the array our algorithm exits without entering for a loop.
So we can check if our functions enter the loop if not we can directly return the value as the answer.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function to find equilibrium point
// a: input array
// n: size of array
int equilibriumPoint(int a[], int n)
{
    // Here we define two pointers to the array -> start =
    // 0, end = n-1 Two variables to take care of sum ->
    // left_sum = 0, right_sum = 0
    int i = 0, start = 0, end = n - 1, left_sum = 0,
        right_sum = 0;
 
    for (i = 0; i < n; i++) {
 
        // if the equilibrium element is found our start
        // will be equal to end variable and  left_sum will
        // be equal right_sum => return the equilibrium
        // element
        if (start == end && right_sum == left_sum)
            return a[start];
 
        // if start is equal to end variable but left_sum is
        // not equal right_sum => no equilibrium element
        // return -1
        if (start == end)
            return -1;
 
        // if left_sum  > right_sum => add the current end
        // element to the right_sum and decrement end
        if (left_sum > right_sum) {
            right_sum += a[end];
            end--;
        }
 
        // if right_sum < left_sum => add the current start
        // element to the left_sum and increment start
        else if (right_sum > left_sum) {
            left_sum += a[start];
            start++;
        }
        /*
            if  left_sum is equal right_sum but start is not
           equal to end => we are still in the middle of
           algorithm even though we found that left_sum is
           equal right_sum we haven't got that one required
           equilibrium element (So, in this case add the
           current end element to the right_sum and
           decrement end (or) add the current start element
           to the left_sum and increment start, to make our
           algorithm continue further)
        */
 
        else {
            right_sum += a[end];
            end--;
        }
    }
 
    // When there is only one element in array our algorithm
    // exits without entering for loop So we can check if our
    // functions enters the loop if not we can directly
    // return the value as answer
    if (!i) {
        return a[0];
    }
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 4, 1, 4, 5 };
    int size = sizeof(arr) / sizeof(arr[0]);
    cout << (equilibriumPoint(arr, size));
}

Java

/*package whatever //do not write package name here */
import java.io.*;
class GFG
{
 
  // Function to find equilibrium point
  // a: input array
  // n: size of array
  static int equilibriumPoint(int a[], int n)
  {
 
    // Here we define two pointers to the array -> start =
    // 0, end = n-1 Two variables to take care of sum ->
    // left_sum = 0, right_sum = 0
    int i = 0, start = 0, end = n - 1, left_sum = 0,
    right_sum = 0;
 
    for (i = 0; i < n; i++)
    {
 
      // if the equilibrium element is found our start
      // will be equal to end variable and  left_sum will
      // be equal right_sum => return the equilibrium
      // element
      if (start == end && right_sum == left_sum)
        return a[start];
 
      // if start is equal to end variable but left_sum is
      // not equal right_sum => no equilibrium element
      // return -1
      if (start == end)
        return -1;
 
      // if left_sum  > right_sum => add the current end
      // element to the right_sum and decrement end
      if (left_sum > right_sum)
      {
        right_sum += a[end];
        end--;
      }
 
      // if right_sum < left_sum => add the current start
      // element to the left_sum and increment start
      else if (right_sum > left_sum)
      {
        left_sum += a[start];
        start++;
      }
      /*
                if  left_sum is equal right_sum but start is not
               equal to end => we are still in the middle of
               algorithm even though we found that left_sum is
               equal right_sum we haven't got that one required
               equilibrium element (So, in this case add the
               current end element to the right_sum and
               decrement end (or) add the current start element
               to the left_sum and increment start, to make our
               algorithm continue further)
            */
 
      else {
        right_sum += a[end];
        end--;
      }
    }
 
    // When there is only one element in array our algorithm
    // exits without entering for loop So we can check if our
    // functions enters the loop if not we can directly
    // return the value as answer
    if (i == 0)
    {
      return a[0];
    }
    return -1;
  }
 
  // Driver code
  public static void main (String[] args)
  {
    int arr[] = { 2, 3, 4, 1, 4, 5 };
    int size = arr.length;
    System.out.println(equilibriumPoint(arr, size));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

<script>
 
// Function to find equilibrium point
// a: input array
// n: size of array
function equilibriumPoint(a, n)
{
    // Here we define two pointers to the array -> start =
    // 0, end = n-1 Two variables to take care of sum ->
    // left_sum = 0, right_sum = 0
    let i = 0, start = 0, end = n - 1, left_sum = 0,
        right_sum = 0;
 
    for (i = 0; i < n; i++) {
 
        // if the equilibrium element is found our start
        // will be equal to end variable and left_sum will
        // be equal right_sum => return the equilibrium
        // element
        if (start == end && right_sum == left_sum)
            return a[start];
 
        // if start is equal to end variable but left_sum is
        // not equal right_sum => no equilibrium element
        // return -1
        if (start == end)
            return -1;
 
        // if left_sum > right_sum => add the current end
        // element to the right_sum and decrement end
        if (left_sum > right_sum) {
            right_sum += a[end];
            end--;
        }
 
        // if right_sum < left_sum => add the current start
        // element to the left_sum and increment start
        else if (right_sum > left_sum) {
            left_sum += a[start];
            start++;
        }
        /*
            if left_sum is equal right_sum but start is not
        equal to end => we are still in the middle of
        algorithm even though we found that left_sum is
        equal right_sum we haven't got that one required
        equilibrium element (So, in this case add the
        current end element to the right_sum and
        decrement end (or) add the current start element
        to the left_sum and increment start, to make our
        algorithm continue further)
        */
 
        else {
            right_sum += a[end];
            end--;
        }
    }
 
    // When there is only one element in array our algorithm
    // exits without entering for loop So we can check if our
    // functions enters the loop if not we can directly
    // return the value as answer
    if (!i) {
        return a[0];
    }
}
 
// Driver code
    let arr = [ 2, 3, 4, 1, 4, 5 ];
    let size = arr.length;
    document.write(equilibriumPoint(arr, size));
 
 
 
 
// This code is contributed by Surbhi Tyagi.
</script>

Output

Time complexity: O(n)

Space complexity: O(1)

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