454. 4Sum II

Medium

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

from collections import defaultdict
class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        # n^4 when doing burte force so lets break that into n^2 + n^2 :D
        d = defaultdict(int)
        for i in nums3:
            for j in nums4:
                d[i+j] += 1
        print(d)
        count = 0
        for value in nums1:
            for val in nums2:
                temp = -(value+val)
                if temp in d and d[temp] != 0:
                    count += d[temp]
        return count