4Sum_II.swift

/*
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 1:
build AB_sum map
- key: sum of A[i], B[j]
- value: number of how many pairs have this sum

for C[i], D[j], find if AB_sum[-i-j] exist, 
if exist, add it's value into counter

Time Complexity: O(n^2)
*/
class Solution {
    func fourSumCount(_ A: [Int], _ B: [Int], _ C: [Int], _ D: [Int]) -> Int {        
        // key is sum, 
        // value is count
        var AB_sum: [Int: Int] = [:]
        A.forEach { i in
            B.forEach { j in
                AB_sum[i+j, default: 0] += 1
            }
        }
        
        var res: Int = 0
        C.forEach { i in
            D.forEach { j in
                res += AB_sum[-i-j, default: 0]
            }
        }
        return res
    }
}