4Sum II.java

/*
	Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

	To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2^{28} to 2^{28} - 1 and the result is guaranteed to be at most 2^{31} - 1.
    
    Link: https://leetcode.com/problems/4sum-ii/

    Example: 

		Input:
			A = [ 1, 2]
			B = [-2,-1]
			C = [-1, 2]
			D = [ 0, 2]

		Output:
			2

		Explanation:
			The two tuples are:
			1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
			2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

    Solution: None

    Source: https://redmine.adelphi.digital/projects/frdc/wiki/Server_details
*/

public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
       	Map<Integer,Integer> sums = new HashMap<>();
		int count = 0;
		for (int i = 0; i < A.length; i++) {
			for (int j = 0;j < B.length; j++){
				int sum = A[i] + B[j];
				if (sums.containsKey(sum)) {
					sums.put(sum, sums.get(sum) + 1);
				} else {
					sums.put(sum, 1);
				}
			}
		}
		for (int k = 0; k < A.length; k++) {
			for (int z = 0; z < C.length; z++){
				int sum = -(C[k] + D[z]);
				if (sums.containsKey(sum)) {
					count += sums.get(sum);
				}
			}
		}
		return count; 
    }
}