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

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:

class Solution { public int solution(int[] 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).

Elements of input arrays can be modified.

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Solution:

import java.math.BigInteger;

public class Solution {

    private static BigInteger sub = BigInteger.ZERO;
    private static BigInteger sum = BigInteger.ZERO;
    private static BigInteger left = BigInteger.ZERO;
    private static BigInteger right = BigInteger.ZERO;

    public static void main(String[] args) throws Exception {
        Solution s = new Solution();

        int[] A = {-1, 3, -4, 5, 1, -6, 2, 1};
        System.out.print(s.solution(A));

    }

    public int solution(int[] A) {

        sum = BigInteger.ZERO;
        if (A.length != -1 ) {
            sum = sum.add(sumOfSubArray(A));
        }

        for (int i=0; i<A.length; i++) {
            sub = sub.add(BigInteger.valueOf(A[i]));

            left = sub.subtract(BigInteger.valueOf(A[i]));
            right = sum.subtract(sub);

            if ( left.equals(right)) {
                return i;
            }
        }

        return -1;
    }

    public static BigInteger sumOfSubArray(int[] a) {
        BigInteger sum = BigInteger.ZERO;

        for (int i = 0; i < a.length; i++) {
            sum = sum.add(BigInteger.valueOf(a[i]));
        }

        return sum;
    }
}