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C Program for Equilibrium index of an array

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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. 
The 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

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 <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;
}


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 <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;
}


Output

First equilibrium index is 3

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

Please refer complete article on Equilibrium index of an array for more details!



Last Updated : 05 Jan, 2023
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