4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples(i, j, k, l)there are such thatA[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 -228 to 228- 1 and the result is guaranteed to be at most 231- 1.

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:

Compute every combination sums of first two array, and then the other two.

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    HashMap<Integer, Integer> map = 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];
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
    }
    for (int i = 0; i < C.length; i++) {
        for (int j = 0; j < D.length; j++) {
            int sum = - (C[i] + D[j]);
            if(map.containsKey(sum)) {
                count += map.get(sum);
            }
        }
    }
    return count;
}