4sum-ii.py

# Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:
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# 	0 <= i, j, k, l < n
# 	nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0
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# Example 1:
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# 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
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# Example 2:
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# Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
# Output: 1
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# Constraints:
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# 	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
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class Solution:
    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
        d1, d2, res = {}, {}, 0
        print(A, B)
        for a, b in product(A, B):
            d1[a+b] = d1.get(a+b, 0)+1
        for c, d in product(C, D):
            d2[c+d] = d2.get(c+d, 0)+1
        for key, value in d1.items():
            # We need to void it to == 0
            if -key in d2:
                
                res += d1[key]*d2[-key]
        return res