Equid: Find an index in an array such that its prefix sum equals its suffix sum.

solution.js

```
This is a demo task.

A zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e. 
A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].
Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.

For example, consider the following array A consisting of N = 8 elements:

  A[0] = -1
  A[1] =  3
  A[2] = -4
  A[3] =  5
  A[4] =  1
  A[5] = -6
  A[6] =  2
  A[7] =  1
P = 1 is an equilibrium index of this array, because:

A[0] = −1 = A[2] + A[3] + A[4] + A[5] + A[6] + A[7]
P = 3 is an equilibrium index of this array, because:

A[0] + A[1] + A[2] = −2 = A[4] + A[5] + A[6] + A[7]
P = 7 is also an equilibrium index, because:

A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] = 0
and there are no elements with indices greater than 7.

P = 8 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.

Write a function:

def solution(a)

that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.

For example, given array A shown above, the function may return 1, 3 or 7, as explained above.

Assume that:

N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
```

var array = [-1, 3, -4, 5, 1, -6, 2, 1];
function solution(A) {
var lSum = 0,
  solution = -1,
  rSum = 0;
  if (A.length > 0) {
    rSum = A.reduce(function(a,b){
      return a + b;
    });
  }
  A.forEach(function(element, index) {
    if (isNaN(element)) {
      return;
    }
    rSum -= element;

    if (lSum === rSum) {
      solution = index;
    }
    lSum += element;
  });
  return solution;
}
var s = solution(array);

This solution took me longer than 30 mins, but finally made it. Also challenged me to search a little in order to find the best solution.