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Problem Statement:

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

Solution:

Create a hashmap H of all possible sums of nums1 and nums2 in O(N^2) time. Then traverse nums3Xnums4 in O(N^2) time and each time check membership in H. Easy-peasy!

C++ version:

 
class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int n = nums1.size(), res=0;
        unordered_map<int,int> H;
        for (int n1: nums1) for (int n2: nums2) H[n1+n2]++;
        for (int n3: nums3) for (int n4: nums4) if (H.count(-(n3+n4))) res+=H[-(n3+n4)];
        return res;
    }
};

Python version:

 
class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        H = collections.defaultdict(int)
        res = 0
        for n1 in nums1:
            for n2 in nums2:
                H[n1+n2]+=1
        for n3 in nums3:
            for n4 in nums4:
                if -(n3+n4) in H:
                    res += H[-(n3+n4)]
        return res