# 4Sum II

### 描述​

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 $-2^{28}$ to $2^{28}$ - 1 and the result is guaranteed to be at most $2^{31}$ - 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

### 分析​

1. 4 层循环暴力搜索，时间复杂度 O($n^4$)，空间复杂度 O(1).
2. 用一个 HashMap 缓存数组 A+B，另一个 HashMap 缓存 C+D，key 为和，value 为和出现的次数，然后用 2 层循环，遍历两个 HashMap，统计和为 0 的次数。该方法时间复杂度 O($n^2$)，空间复杂度 O($n^2$)，用空间换取了时间。
3. 推广到 k 个数组，也可以用上述方法，用一个 HashMap 缓存前$\frac{k}{2}$个数组的和，另一个 HashMap 缓存后$\frac{k}{2}$个数组的和，然后遍历两个 HashMap，统计和为 0 的次数。时间复杂度 O($n^{\frac{k}{2}}$)，空间复杂度 O($n^{\frac{k}{2}}$)。

### 代码​

# 4Sum II# HashMap缓存和出现的次数# Time Complexity: O(n^2)，Space Complexity: O(n^2)from collections import Counterfrom itertools import productclass Solution:    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:        AB = Counter(a+b for a, b in product(A, B))        CD = Counter(c+d for c, d in product(C, D))        count = 0        for num in AB:            if -num in CD:                count += AB[num] * CD[-num]        return count