454. 4Sum II

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
  • One HashMap
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    HashMap<Integer, Integer> map = new HashMap<>();
    HashMap<Integer, Integer> map2 = new HashMap<>();
    for(int a : A) {
        for(int b : B)
            map.put(a+b, map.getOrDefault(a+b, 0)+1);
    }
    int result = 0;
    for(int c : C) {
        for(int d : D) {
            if(map.containsKey(-(c+d))) {
                result += map.get(-(c+d));
            }
        }
    }
    return result;
}
  • Two HashMap
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    HashMap<Integer, Integer> map = new HashMap<>();
    HashMap<Integer, Integer> map2 = new HashMap<>();
    for(int a : A) {
        for(int b : B)
            map.put(a+b, map.getOrDefault(a+b, 0)+1);
    }
    for(int a : C) {
        for(int b : D)
            map2.put(a+b, map2.getOrDefault(a+b, 0)+1);
    }
    int result = 0;
    for(int key : map2.keySet()) {
        if(map.containsKey(-key)) {
            result += map.get(-key) * map2.get(key);
        }
    }
    return result;
}