Sum of the 4 (2)

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 -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

Complexity:

• Time complexity: O(n^2)
• Space complexity: O(1)

function fourSumCount(A, B, C, D) {
    const map = {};
    let res = 0;
    let key = 0;

    for (let i = 0; i < A.length; i++) {
        for (let j = 0; j < B.length; j++) {
            key = A[i] + B[j];
            map[key] = (map[key] || 0) + 1;
        }
    }

    for (let i = 0; i < C.length; i++) {
        for (let j = 0; j < D.length; j++) {
            key = -(C[i] + D[j]);
            res += map[key] || 0;
        }
    }

    return res;
}