[FIXED] A zero-indexed array given & An equilibrium index of this array

Issue

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.

Now i have to write a function:

int solution(int A[], int N);

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].

here have some Complexity:

Elements of input arrays can be modified.

Solution

100 Score in Javascript

function solution(V) {

    var sum = 0;
    for (i=0; i < V.length; i++) {
      sum += V[i];   
    }

    var leftSum= 0;
    var rightSum = 0;

    for (j=0; j < V.length; j++) {
      rightSum = sum - (leftSum + V[j]);
      if(leftSum == rightSum) {
          return j;
      }
      leftSum += V[j];
    }
    return -1;
}

Answered By - Eko Bayu
Answer Checked By - Marie Seifert (JavaFixing Admin)