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Equilibrium index of an array

January 7, 2011

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 = -7, A = 1, A = 5, A = 2, A = -4, A = 3, A=0

3 is an equilibrium index, because:
A + A + A = A + A + A

6 is also an equilibrium index, because sum of zero elements is zero, i.e., A + A + A + A + A + A=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).

 `#include ` `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)``  ``{``    ``leftsum = 0;  ``// initialize left sum for current index i``    ``rightsum = 0; ``// initialize right sum for current index i` `    ``/* 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;``}` `int` `main()``{``  ``int` `arr[] = {-7, 1, 5, 2, -4, 3, 0};``  ``int` `arr_size = ``sizeof``(arr)/``sizeof``(arr);``  ``printf``(``"%d\n"``, equilibrium(arr, arr_size));` `  ``getchar``();``  ``return` `0;``}`

Time Complexity: 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.

```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 ` `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);``  ``printf``(``"First equilibrium index is %d\n"``, equilibrium(arr, arr_size));` `  ``getchar``();``  ``return` `0;``}`

Time Complexity: O(n)

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.