[EAZY] - TapeEquilibrium

Task description A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape. Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1]. The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])| In other words, it is the absolute difference between the sum of the first part and the sum of the second part. For example, consider array A such that: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 We can split this tape in four places: P = 1, difference = |3 − 10| = 7  P = 2, difference = |4 − 9| = 5  P = 3, difference = |6 − 7| = 1  P = 4, difference = |10 − 3| = 7  Write a function: class Solution { public int solution(int[] A); } that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved. For example, given: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 the function should return 1, as explained above. Assume that: N is an integer within the range [2..100,000]; each element of array A is an integer within the range [−1,000..1,000]. 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 modifie

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        
        if (A == null || A.length < 2 || A.length > 100000) {
		return 0;
	}

	if (A.length == 2) {
	        return Math.abs(A[0] - A[1]);
	}

	int leftSum = A[0];
	int rightSum = 0;

	for (int j=1; j<A.length; j++) {
		rightSum += A[j];
	}

	int minDiff = Math.abs(leftSum - rightSum);

	for (int i=1; i<A.length-1; i++) {
		leftSum += A[i];
		rightSum -= A[i];
		int diff = Math.abs(leftSum - rightSum);
		if (diff < minDiff) {
			minDiff = diff;
		}
	}

	return minDiff;        
    }
}