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  • Difficulty Level : Easy
  • Last Updated : 13 Jul, 2022

Equilibrium index of an array is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes. For example, in an array A: 

Example : 

Input: A[] = {-7, 1, 5, 2, -4, 3, 0} 
Output: 3 
3 is an equilibrium index, because: 
A[0] + A[1] + A[2] = A[4] + A[5] + A[6]

Input: A[] = {1, 2, 3} 
Output: -1 

Write a function int equilibrium(int[] arr, int n); that given a sequence arr[] of size n, returns an equilibrium index (if any) or -1 if no equilibrium indexes exist. 

Recommended Practice

Method 1 (Simple but inefficient) 
Use two loops. Outer loop iterates through all the element and inner loop finds out whether the current index picked by the outer loop is equilibrium index or not. Time complexity of this solution is O(n^2). 

C++




// C++ program to find equilibrium
// index of an array
#include <bits/stdc++.h>
using namespace std;
  
int equilibrium(int arr[], int n)
{
    int i, j;
    int leftsum, rightsum;
  
    /* Check for indexes one by one until 
    an equilibrium index is found */
    for (i = 0; i < n; ++i) 
    {     
  
        /* get left sum */
        leftsum = 0; 
        for (j = 0; j < i; j++)
            leftsum += arr[j];
  
        /* get right sum */
        rightsum = 0; 
        for (j = i + 1; j < n; j++)
            rightsum += arr[j];
  
        /* if leftsum and rightsum  
        are same, then we are done */
        if (leftsum == rightsum)
            return i;
    }
  
    /* return -1 if no equilibrium 
    index is found */
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    cout << equilibrium(arr, arr_size);
    return 0;
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)


C




// C program to find equilibrium
// index of an array
  
#include <stdio.h>
  
int equilibrium(int arr[], int n)
{
    int i, j;
    int leftsum, rightsum;
  
    /* Check for indexes one by one until 
      an equilibrium index is found */
    for (i = 0; i < n; ++i) {       
  
        /* get left sum */
        leftsum = 0; 
        for (j = 0; j < i; j++)
            leftsum += arr[j];
  
        /* get right sum */
        rightsum = 0; 
        for (j = i + 1; j < n; j++)
            rightsum += arr[j];
  
        /* if leftsum and rightsum are same, 
           then we are done */
        if (leftsum == rightsum)
            return i;
    }
  
    /* return -1 if no equilibrium index is found */
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    printf("%d", equilibrium(arr, arr_size));
  
    getchar();
    return 0;
}


Java




// Java program to find equilibrium
// index of an array
  
class EquilibriumIndex {
    int equilibrium(int arr[], int n)
    {
        int i, j;
        int leftsum, rightsum;
  
        /* Check for indexes one by one until 
           an equilibrium index is found */
        for (i = 0; i < n; ++i) {
  
            /* get left sum */
            leftsum = 0;  
            for (j = 0; j < i; j++)
                leftsum += arr[j];
  
            /* get right sum */
            rightsum = 0;
            for (j = i + 1; j < n; j++)
                rightsum += arr[j];
  
            /* if leftsum and rightsum are same, 
               then we are done */
            if (leftsum == rightsum)
                return i;
        }
  
        /* return -1 if no equilibrium index is found */
        return -1;
    }
    // Driver code
    public static void main(String[] args)
    {
        EquilibriumIndex equi = new EquilibriumIndex();
        int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
        int arr_size = arr.length;
        System.out.println(equi.equilibrium(arr, arr_size));
    }
}
  
// This code has been contributed by Mayank Jaiswal


Python3




# Python program to find equilibrium 
# index of an array
  
# function to find the equilibrium index
def equilibrium(arr):
    leftsum = 0
    rightsum = 0
    n = len(arr)
  
    # Check for indexes one by one 
    # until an equilibrium index is found
    for i in range(n):
        leftsum = 0
        rightsum = 0
      
        # get left sum
        for j in range(i):
            leftsum += arr[j]
          
        # get right sum
        for j in range(i + 1, n):
            rightsum += arr[j]
          
        # if leftsum and rightsum are same,
        # then we are done
        if leftsum == rightsum:
            return i
      
    # return -1 if no equilibrium index is found
    return -1
              
# driver code
arr = [-7, 1, 5, 2, -4, 3, 0]
print (equilibrium(arr))
  
# This code is contributed by Abhishek Sharama


C#




// C# program to find equilibrium
// index of an array
  
using System;
  
class GFG {
    static int equilibrium(int[] arr, int n)
    {
        int i, j;
        int leftsum, rightsum;
  
        /* Check for indexes one by 
         one until an equilibrium
        index is found */
        for (i = 0; i < n; ++i) {
  
            // initialize left sum
            // for current index i
            leftsum = 0;
  
            // initialize right sum
            // for current index i
            rightsum = 0;
  
            /* get left sum */
            for (j = 0; j < i; j++)
                leftsum += arr[j];
  
            /* get right sum */
            for (j = i + 1; j < n; j++)
                rightsum += arr[j];
  
            /* if leftsum and rightsum are
             same, then we are done */
            if (leftsum == rightsum)
                return i;
        }
  
        /* return -1 if no equilibrium 
         index is found */
        return -1;
    }
  
    // driver code
    public static void Main()
    {
        int[] arr = { -7, 1, 5, 2, -4, 3, 0 };
        int arr_size = arr.Length;
  
        Console.Write(equilibrium(arr, arr_size));
    }
}
  
// This code is contributed by Sam007


PHP




<?php 
// PHP program to find equilibrium 
// index of an array 
  
function equilibrium($arr, $n
    $i; $j
    $leftsum;
    $rightsum
  
    /* Check for indexes one by one until 
    an equilibrium index is found */
    for ($i = 0; $i < $n; ++$i
    {     
  
        /* get left sum */
        $leftsum = 0; 
        for ($j = 0; $j < $i; $j++) 
            $leftsum += $arr[$j]; 
  
        /* get right sum */
        $rightsum = 0; 
        for ($j = $i + 1; $j < $n; $j++) 
            $rightsum += $arr[$j]; 
  
        /* if leftsum and rightsum 
        are same, then we are done */
        if ($leftsum == $rightsum
            return $i
    
  
    /* return -1 if no equilibrium
       index is found */
    return -1; 
  
// Driver code 
$arr = array( -7, 1, 5, 2, -4, 3, 0 ); 
$arr_size = sizeof($arr); 
echo equilibrium($arr, $arr_size); 
  
// This code is contributed 
// by akt_mit
?>


Javascript




<script>
// JavaScript Program to find equilibrium
// index of an array
function equilibrium(arr, n)
{
         var i, j;
         var leftsum, rightsum;
           
         /*Check for indexes one by one until 
         an equilibrium index is found*/
         for(i = 0; i < n; ++i)
         {
           
             /*get left sum*/
             leftsum = 0;
              for(let j = 0; j < i; j++)
              leftsum += arr[j];
                
              /*get right sum*/
              rightsum = 0;
              for(let j = i + 1; j < n; j++)
              rightsum += arr[j];
                
              /*if leftsum and rightsum are same,
              then we are done*/
              if(leftsum == rightsum)
                 return i;
         }
           
         /* return -1 if no equilibrium index is found*/
            return -1;
}
     // Driver code
       
     var arr = new Array(-7,1,5,2,-4,3,0);
     n = arr.length;
      document.write(equilibrium(arr,n));
        
// This code is contributed by simranarora5sos  
</script>


Output

3

Time Complexity: O(n^2)

Auxiliary Space: O(1)

Method 2 (Tricky and Efficient) 
The idea is to get the total sum of the array first. Then Iterate through the array and keep updating the left sum which is initialized as zero. In the loop, we can get the right sum by subtracting the elements one by one. Thanks to Sambasiva for suggesting this solution and providing code for this.

1) Initialize leftsum  as 0
2) Get the total sum of the array as sum
3) Iterate through the array and for each index i, do following.
    a)  Update sum to get the right sum.  
           sum = sum - arr[i] 
       // sum is now right sum
    b) If leftsum is equal to sum, then return current index. 
       // update leftsum for next iteration.
    c) leftsum = leftsum + arr[i]
4) return -1 
// If we come out of loop without returning then
// there is no equilibrium index

The image below shows the dry run of the above approach: 

Below is the implementation of the above approach: 

C++




// C++ program to find equilibrium 
// index of an array 
#include <bits/stdc++.h>
using namespace std;
  
int equilibrium(int arr[], int n) 
    int sum = 0; // initialize sum of whole array 
    int leftsum = 0; // initialize leftsum 
  
    /* Find sum of the whole array */
    for (int i = 0; i < n; ++i) 
        sum += arr[i]; 
  
    for (int i = 0; i < n; ++i) 
    
        sum -= arr[i]; // sum is now right sum for index i 
  
        if (leftsum == sum) 
            return i; 
  
        leftsum += arr[i]; 
    
  
    /* If no equilibrium index found, then return 0 */
    return -1; 
  
// Driver code 
int main() 
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 }; 
    int arr_size = sizeof(arr) / sizeof(arr[0]); 
    cout << "First equilibrium index is " << equilibrium(arr, arr_size); 
    return 0; 
  
// This is code is contributed by rathbhupendra


C




// C program to find equilibrium
// index of an array
  
#include <stdio.h>
  
int equilibrium(int arr[], int n)
{
    int sum = 0; // initialize sum of whole array
    int leftsum = 0; // initialize leftsum
  
    /* Find sum of the whole array */
    for (int i = 0; i < n; ++i)
        sum += arr[i];
  
    for (int i = 0; i < n; ++i) {
        sum -= arr[i]; // sum is now right sum for index i
  
        if (leftsum == sum)
            return i;
  
        leftsum += arr[i];
    }
  
    /* If no equilibrium index found, then return 0 */
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    printf("First equilibrium index is %d"
                 equilibrium(arr, arr_size));
  
    getchar();
    return 0;
}


Java




// Java program to find equilibrium
// index of an array
  
class EquilibriumIndex {
    int equilibrium(int arr[], int n)
    {
        int sum = 0; // initialize sum of whole array
        int leftsum = 0; // initialize leftsum
  
        /* Find sum of the whole array */
        for (int i = 0; i < n; ++i)
            sum += arr[i];
  
        for (int i = 0; i < n; ++i) {
            sum -= arr[i]; // sum is now right sum for index i
  
            if (leftsum == sum)
                return i;
  
            leftsum += arr[i];
        }
  
        /* If no equilibrium index found, then return 0 */
        return -1;
    }
  
   // Driver code
    public static void main(String[] args)
    {
        EquilibriumIndex equi = new EquilibriumIndex();
        int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
        int arr_size = arr.length;
        System.out.println("First equilibrium index is "
                          equi.equilibrium(arr, arr_size));
    }
}
  
// This code has been contributed by Mayank Jaiswal


Python3




# Python program to find the equilibrium
# index of an array
  
# function to find the equilibrium index
def equilibrium(arr):
  
    # finding the sum of whole array
    total_sum = sum(arr)
    leftsum = 0
    for i, num in enumerate(arr):
          
        # total_sum is now right sum
        # for index i
        total_sum -= num
          
        if leftsum == total_sum:
            return i
        leftsum += num
       
      # If no equilibrium index found, 
      # then return -1
    return -1
      
# Driver code
arr = [-7, 1, 5, 2, -4, 3, 0]
print ('First equilibrium index is ',
       equilibrium(arr))
  
# This code is contributed by Abhishek Sharma


C#




// C# program to find the equilibrium
// index of an array
  
using System;
  
class GFG {
    static int equilibrium(int[] arr, int n)
    {
        // initialize sum of whole array
        int sum = 0;
  
        // initialize leftsum
        int leftsum = 0;
  
        /* Find sum of the whole array */
        for (int i = 0; i < n; ++i)
            sum += arr[i];
  
        for (int i = 0; i < n; ++i) {
  
            // sum is now right sum
            // for index i
            sum -= arr[i];
  
            if (leftsum == sum)
                return i;
  
            leftsum += arr[i];
        }
  
        /* If no equilibrium index found, 
        then return 0 */
        return -1;
    }
  
    // Driver code
    public static void Main()
    {
        int[] arr = { -7, 1, 5, 2, -4, 3, 0 };
        int arr_size = arr.Length;
  
        Console.Write("First equilibrium index is " +
                        equilibrium(arr, arr_size));
    }
}
// This code is contributed by Sam007


PHP




<?php
// PHP program to find equilibrium 
// index of an array 
  
function equilibrium($arr, $n
    $sum = 0; // initialize sum of 
              // whole array 
    $leftsum = 0; // initialize leftsum 
  
    /* Find sum of the whole array */
    for ($i = 0; $i < $n; ++$i
        $sum += $arr[$i]; 
  
    for ($i = 0; $i < $n; ++$i
    
        // sum is now right sum 
        // for index i 
        $sum -= $arr[$i]; 
  
        if ($leftsum == $sum
            return $i
  
        $leftsum += $arr[$i]; 
    
  
    /* If no equilibrium index 
    found, then return 0 */
    return -1; 
  
// Driver code
$arr = array( -7, 1, 5, 2, -4, 3, 0 ); 
$arr_size = sizeof($arr); 
echo "First equilibrium index is "
      equilibrium($arr, $arr_size); 
  
// This code is contributed by ajit
?>


Javascript




<script>
// program to find equilibrium
// index of an array
 function equilibrium(arr, n)
{
    sum = 0; // initialize sum of whole array
    leftsum = 0; // initialize leftsum
   
    /* Find sum of the whole array */
    for (let i = 0; i < n; ++i)
        sum += arr[i];
   
    for (let i = 0; i < n; ++i)
    {
        sum -= arr[i]; // sum is now right sum for index i
   
        if (leftsum == sum)
            return i;
   
        leftsum += arr[i];
    }
   
    /* If no equilibrium index found, then return 0 */
    return -1;
}
   
// Driver code
  
arr =new Array(-7, 1, 5, 2, -4, 3, 0);
n=arr.length;
document.write("First equilibrium index is " + equilibrium(arr, n));
  
// This code is contributed by simranarora5sos
</script>


Output

First equilibrium index is 3

 Time Complexity: O(n)

Auxiliary Space: O(1)

Method 3 :

This is a quite simple and straightforward method. The idea is to take the prefix sum of the array twice. Once from the front end of array and another from the back end of array.

After taking both prefix sums run a loop and check for some i if both the prefix sum from one array is equal to prefix sum from the second array then that point can be considered as the Equilibrium point.

C++




// C++ program to find equilibrium index of an array
#include <bits/stdc++.h>
using namespace std;
  
int equilibrium(int a[], int n)
{
    if (n == 1)
        return (0);
    int forward[n] = { 0 };
    int rev[n] = { 0 };
  
    // Taking the prefixsum from front end array
    for (int i = 0; i < n; i++) {
        if (i) {
            forward[i] = forward[i - 1] + a[i];
        }
        else {
            forward[i] = a[i];
        }
    }
  
    // Taking the prefixsum from back end of array
    for (int i = n - 1; i > 0; i--) {
        if (i <= n - 2) {
            rev[i] = rev[i + 1] + a[i];
        }
        else {
            rev[i] = a[i];
        }
    }
  
    // Checking if forward prefix sum
    // is equal to rev prefix
    // sum
    for (int i = 0; i < n; i++) {
        if (forward[i] == rev[i]) {
            return i;
        }
    }
    return -1;
  
    // If You want all the points
    // of equilibrium create
    // vector and push all equilibrium
    // points in it and
    // return the vector
}
  
// Driver code
int main()
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "First Point of equilibrium is at index "
         << equilibrium(arr, n) << "\n";
    return 0;
}


Java




// Java program to find equilibrium
// index of an array
class GFG{
  
static int equilibrium(int a[], int n)
    if (n == 1)
        return (0);
      
    int[] front = new int[n];
    int[] back = new int[n];
  
    // Taking the prefixsum from front end array
    for (int i = 0; i < n; i++)
    {
        if (i != 0)
        {
            front[i] = front[i - 1] + a[i];
        }
        else 
        {
            front[i] = a[i];
        }
    }
    
    // Taking the prefixsum from back end of array
    for (int i = n - 1; i > 0; i--) 
    {
        if (i <= n - 2
        {
            back[i] = back[i + 1] + a[i];
        }
        else 
        {
            back[i] = a[i];
        }
    }
      
    // Checking for equilibrium index by
    //comparing front and back sums 
    for(int i = 0; i < n; i++) 
    {
        if (front[i] == back[i])
        {
            return i;
        }
    }
      
    // If no equilibrium index found,then return -1
    return -1;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = arr.length;
      
    System.out.println("First Point of equilibrium " +
                       "is at index "
                       equilibrium(arr, arr_size));
}
}
  
// This code is contributed by Lovish Aggarwal


Python3




# Python program to find the equilibrium
# index of an array
  
# Function to find the equilibrium index
def equilibrium(arr):
    left_sum = []
    right_sum = []
  
    # Iterate from 0 to len(arr)
    for i in range(len(arr)):
  
        # If i is not 0
        if(i):
            left_sum.append(left_sum[i-1]+arr[i])
            right_sum.append(right_sum[i-1]+arr[len(arr)-1-i])
        else:
            left_sum.append(arr[i])
            right_sum.append(arr[len(arr)-1])
  
    # Iterate from 0 to len(arr)    
    for i in range(len(arr)):
        if(left_sum[i] == right_sum[len(arr) - 1 - i ]):
            return(i)
            
    # If no equilibrium index found,then return -1
    return -1
  
  
# Driver code
arr = [-7, 1, 5, 2, -4, 3, 0]
print('First equilibrium index is ',
      equilibrium(arr))
  
# This code is contributed by Lokesh Sharma


C#




// C# program to find equilibrium
// index of an array
using System;
  
class GFG{
  
static int equilibrium(int[] a, int n)
{
    if (n == 1)
        return (0);
  
    int[] front = new int[n];
    int[] back = new int[n];
  
    // Taking the prefixsum from front end array
    for(int i = 0; i < n; i++) 
    {
        if (i != 0)
        {
            front[i] = front[i - 1] + a[i];
        }
        else 
        {
            front[i] = a[i];
        }
    }
  
    // Taking the prefixsum from back end of array
    for(int i = n - 1; i > 0; i--)
    {
        if (i <= n - 2) 
        {
            back[i] = back[i + 1] + a[i];
        }
        else
        {
            back[i] = a[i];
        }
    }
  
    // Checking for equilibrium index by
    // comparing front and back sums
    for(int i = 0; i < n; i++) 
    {
        if (front[i] == back[i])
        {
            return i;
        }
    }
  
    // If no equilibrium index found,then return -1
    return -1;
}
  
// Driver code
public static void Main(string[] args)
{
    int[] arr = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = arr.Length;
  
    Console.WriteLine("First Point of equilibrium "
                      "is at index "
                      equilibrium(arr, arr_size));
}
}
  
// This code is contributed by ukasp


Javascript




<script>
// Program to find equilibrium index of an array
 function equilibrium(a, n)
{
    if (n == 1)
        return (0);
    var forward = new Array(0);
    var rev = new Array(0);
   
    // Taking the prefixsum from front end array
    for (let i = 0; i < n; i++) {
        if (i) {
            forward[i] = forward[i - 1] + a[i];
        }
        else {
            forward[i] = a[i];
        }
    }
   
    // Taking the prefixsum from back end of array
    for (let i = n - 1; i > 0; i--) {
        if (i <= n - 2) {
            rev[i] = rev[i + 1] + a[i];
        }
        else {
            rev[i] = a[i];
        }
    }
   
    // Checking if forward prefix sum
    // is equal to rev prefix
    // sum
    for (let i = 0; i < n; i++) {
        if (forward[i] == rev[i]) {
            return i;
        }
    }
    return -1;
   
    // If You want all the points
    // of equilibrium create
    // vector and push all equilibrium
    // points in it and
    // return the vector
}
   
// Driver code
    arr = new Array(-7, 1, 5, 2, -4, 3, 0);
    n = arr.length;
    document.write("First Point of equilibrium is at index "
         + equilibrium(arr, n) + "\n");
           
// This code is contributed by simranarora5sos
</script>


Output

First Point of equilibrium is at index 3

Time Complexity: O(n)

Auxiliary Space : O(n)

Method 4:– Using binary search

To handle all the testcase, we can use binary search algorithm.

1.calculate the mid and then create left sum and right sum around mid

2.if left sum is greater than right sum, move to left until it become equal or less than right sum

3. else if right sum is greater than left, move right until it become equal or less than left sum.

4. finally we compare two sums if they are equal we got mid as index else its -1

C++




#include <bits/stdc++.h>
using namespace std;
  
void find(int arr[],  int n)
{
    int mid = n / 2;
    int leftSum = 0, rightSum = 0;
  
    //calculation sum to left of mid
    for (int i = 0; i < mid; i++)
    {
        leftSum += arr[i];
    }
    //calculating sum to right of mid
    for (int i = n - 1; i > mid; i--)
    {
        rightSum += arr[i];
    }
  
    //if rightsum > leftsum
    if (rightSum > leftSum)
    {
        //we keep moving right until rightSum become equal or less than leftSum
        while (rightSum > leftSum && mid < n - 1)
        {
            rightSum -= arr[mid + 1];
            leftSum += arr[mid];
            mid++;
        }
    }
    else
    {
        //we keep moving right until leftSum become equal or less than RightSum
        while (leftSum > rightSum && mid > 0)
        {
            rightSum += arr[mid];
            leftSum -= arr[mid - 1];
            mid--;
        }
    }
  
    //check if both sum become equal
    if (rightSum == leftSum)
    {
        cout <<"First Point of equilibrium is at index ="<< mid << endl;
        return;
    }
  
    cout <<"First Point of equilibrium is at index ="<< -1 << endl;
}
int main()
{
    int arr[] = { 1,1,1,-1,1,1,1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    find(arr, n);
}


Java




// Java program for the above approach
import java.io.*;
import java.util.*;
  
class GFG{
  
static void find(int arr[],  int n)
{
    int mid = n / 2;
    int leftSum = 0, rightSum = 0;
  
    //calculation sum to left of mid
    for (int i = 0; i < mid; i++)
    {
        leftSum += arr[i];
    }
    //calculating sum to right of mid
    for (int i = n - 1; i > mid; i--)
    {
        rightSum += arr[i];
    }
  
    //if rightsum > leftsum
    if (rightSum > leftSum)
    {
        //we keep moving right until rightSum become equal or less than leftSum
        while (rightSum > leftSum && mid < n - 1)
        {
            rightSum -= arr[mid + 1];
            leftSum += arr[mid];
            mid++;
        }
    }
    else
    {
        //we keep moving right until leftSum become equal or less than RightSum
        while (leftSum > rightSum && mid > 0)
        {
            rightSum += arr[mid];
            leftSum -= arr[mid - 1];
            mid--;
        }
    }
  
    //check if both sum become equal
    if (rightSum == leftSum)
    {
        System.out.print("First Point of equilibrium is at index ="+ mid);
        return;
    }
  
    System.out.print("First Point of equilibrium is at index =" + -1);
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1,1,1,-1,1,1,1 };
    int n = arr.length;
    find(arr, n);
}
}


Python3




# Python program for the above approach
def find(arr, n):
    mid = n // 2;
    leftSum = 0;
    rightSum = 0;
  
    # calculation sum to left of mid
    for i in range(mid):
        leftSum += arr[i];
      
    # calculating sum to right of mid
    for i in range(n - 1, mid, -1):
        rightSum += arr[i];
      
  
    # if rightsum > leftsum
    if (rightSum > leftSum):
        
        # we keep moving right until rightSum become equal or less than leftSum
        while (rightSum > leftSum and mid < n - 1):
            rightSum -= arr[mid + 1];
            leftSum += arr[mid];
            mid += 1;
          
    else:
        # we keep moving right until leftSum become equal or less than RightSum
        while (leftSum > rightSum and mid > 0):
            rightSum += arr[mid];
            leftSum -= arr[mid - 1];
            mid -= 1;
          
    # check if both sum become equal
    if (rightSum == leftSum):
        print("First Point of equilibrium is at index =" , mid);
        return;
    print("First Point of equilibrium is at index =" , -1);
  
# Driver code
if __name__ == '__main__':
    arr = [ 1, 1, 1, -1, 1, 1, 1 ];
    n = len(arr);
    find(arr, n);
  
# This code is contributed by gauravrajput1 


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
  
public class GFG {
      
static void find(int[] arr,  int n)
{
    int mid = n / 2;
    int leftSum = 0, rightSum = 0;
  
    //calculation sum to left of mid
    for (int i = 0; i < mid; i++)
    {
        leftSum += arr[i];
    }
    //calculating sum to right of mid
    for (int i = n - 1; i > mid; i--)
    {
        rightSum += arr[i];
    }
  
    //if rightsum > leftsum
    if (rightSum > leftSum)
    {
        //we keep moving right until rightSum become equal or less than leftSum
        while (rightSum > leftSum && mid < n - 1)
        {
            rightSum -= arr[mid + 1];
            leftSum += arr[mid];
            mid++;
        }
    }
    else
    {
        //we keep moving right until leftSum become equal or less than RightSum
        while (leftSum > rightSum && mid > 0)
        {
            rightSum += arr[mid];
            leftSum -= arr[mid - 1];
            mid--;
        }
    }
  
    //check if both sum become equal
    if (rightSum == leftSum)
    {
        Console.Write("First Point of equilibrium is at index ="+ mid);
        return;
    }
  
    Console.Write("First Point of equilibrium is at index =" + -1);
}
  
  
// Driver Code
public static void Main (string[] args) {
      
    int[] arr = { 1,1,1,-1,1,1,1 };
    int n = arr.Length;
    find(arr, n);
}
}
  
// This code is contributed by code_hunt.


Javascript




<script>
// javascript program for the above approach    
function find(arr , n) {
        var mid = parseInt(n / 2);
        var leftSum = 0, rightSum = 0;
  
        // calculation sum to left of mid
        for (i = 0; i < mid; i++) {
            leftSum += arr[i];
        }
          
        // calculating sum to right of mid
        for (i = n - 1; i > mid; i--) {
            rightSum += arr[i];
        }
  
        // if rightsum > leftsum
        if (rightSum > leftSum)
        {
          
            // we keep moving right until rightSum
            // become equal or less than leftSum
            while (rightSum > leftSum && mid < n - 1) {
                rightSum -= arr[mid + 1];
                leftSum += arr[mid];
                mid++;
            }
        }
        else
        {
            // we keep moving right until leftSum 
            // become equal or less than RightSum
            while (leftSum > rightSum && mid > 0) {
                rightSum += arr[mid];
                leftSum -= arr[mid - 1];
                mid--;
            }
        }
  
        // check if both sum become equal
        if (rightSum == leftSum) {
            document.write("First Point of equilibrium is at index =" + mid);
            return;
        }
  
        document.write("First Point of equilibrium is at index =" + -1);
    }
  
    // Driver code
        var arr = [ 1, 1, 1, -1, 1, 1, 1 ];
        var n = arr.length;
        find(arr, n);
  
// This code is contributed by gauravrajput1 
</script>


Output

First Point of equilibrium is at index =3

Time Complexity: O(n)

Auxiliary Space: O(1)
 

As pointed out by Sameer, we can remove the return statement and add a print statement to print all equilibrium indexes instead of returning only one. 

Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.

 


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