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 = 7

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

because A[0] + A[1] + A[2] = A[4] + A[5] + A[6]. P = 6 is also an equilibrium index, because: A[0] + A[1] + A[2] + A[3] + A[4] + A[5] = 0 and there are no elements with indices greater than 6.

P = 7 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N. Write a function class Solution { public int equi(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. Assume that: N is an integer within the range [0..10,000,000]; each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, given array A such that A[0] = -7 A[1] = 1 A[2] = 5 A[3] = 2 A[4] = -4 A[5] = 3 A[6] = 0 the function may return 3 or 6, as explained above. 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).

**Solution in Java**

// you can also use imports, for example: // import java.math.*; class Solution { public int equi ( int[] A ) { long sum = 0; int i = 0; for (i = 0; i < A.length; i++) { sum += (long) A[i]; } long sum_left = 0; for (i = 0; i < A.length; i++) { long sum_right = sum - sum_left - (long) A[i]; if (sum_left == sum_right) { return i; } sum_left += (long) A[i]; } return -1;// write your code here } }

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