Twitter OA prepare: Equilibrium index of an array

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 arrya A:

A[0] = -7, A[1] = 1, A[2] = 5, A[3] = 2, A[4] = -4, A[5] = 3, A[6]=0

3 is an equilibrium index, because:
A[0] + A[1] + A[2] = A[4] + A[5] + A[6]

6 is also an equilibrium index, because sum of zero elements is zero, i.e., A[0] + A[1] + A[2] + A[3] + A[4] + A[5]=0

7 is not an equilibrium index, because it is not a valid index of array A.

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.

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).

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

O(n) time complexity

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. 
    c) leftsum = leftsum + arr[i] // update leftsum for next iteration.
4) return -1 // If we come out of loop without returning then
             // there is no equilibrium index

#include <stdio.h>
 
int equilibrium(int arr[], int n)
{
   int sum = 0;      // initialize sum of whole array
   int leftsum = 0; // initialize leftsum
   int i;
 
   /* Find sum of the whole array */
   for (i = 0; i < n; ++i)
        sum += arr[i];
 
   for( 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;
}
 
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\n", equilibrium(arr, arr_size));
 
  getchar();
  return 0;
}