動機

這題是靠觀察時間複雜度找到解法的

Problem

Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:

  • 0 <= i, j, k, l < n
  • nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

 

Example 1:

Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]Output: 2Explanation:The two tuples are:1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0

Example 2:

Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]Output: 1

 

Constraints:

  • n == nums1.length
  • n == nums2.length
  • n == nums3.length
  • n == nums4.length
  • 1 <= n <= 200
  • -228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228

Sol

如果硬幹是O(n^4)，但如果當成某種3sum的話，就應該是合理的複雜度，也不會比這個更好，因為是必要試過所有組合

class Solution:
    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
        tmp = []
        for c in C:
            for d in D:
                tmp.append(c+d)
        candidates = Counter(tmp)
        
        ret = 0
        for a in A:
            for b in B:
                if -(a+b) in candidates:
                    ret += candidates[-(a+b)]
        return ret