LC 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

能够优化到O（n2），但是可以只用一个map，如下，我本来的想法是用两个map，但这样就浪费了空间。

Runtime: 155 ms, faster than 67.41% of Java online submissions for 4Sum II.

class Solution {
  private Map<Integer, Integer> mp1 = new HashMap<>();
  public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    for(int i = ; i<A.length; i++){
      for(int j=; j<B.length; j++){
        mp1.put(A[i] + B[j], mp1.getOrDefault(A[i]+B[j],) + );
      }
    }
    int ret = ;
    for(int i = ; i<C.length; i++){
      for(int j=; j<D.length; j++){
        ret += mp1.getOrDefault(-*(C[i]+D[j]), );
      }
    }
    return ret;
  }
}