4Sum II LeetCode Solution

Problem – 4Sum II

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

4Sum II LeetCode Solution in Java

public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
	var pairCountBySum = new HashMap<Integer, Integer>();
	
	Arrays.stream(nums1)
		  .forEach(num1 -> Arrays.stream(nums2)
							     .forEach(num2 -> pairCountBySum.compute(num1 + num2, (k, sumCount) -> sumCount == null ? 1 : ++sumCount)));
		  
	return Arrays.stream(nums3)
				 .map(num3 -> Arrays.stream(nums4)
								    .map(num4 -> pairCountBySum.getOrDefault(-(num3 + num4), 0))
									.sum())
				 .sum();
}

4Sum II LeetCode Solution in Python

def fourSumCount(self, A, B, C, D):
    AB = collections.Counter(a+b for a in A for b in B)
    return sum(AB[-c-d] for c in C for d in D)

4Sum II LeetCode Solution in C++

    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int res = 0;
        unordered_map<int, int> AB;
        for (int a : A)
            for (int b : B)
                AB[a + b]++;
        for (int c : C)
            for (int d : D)
                res += AB[-c - d];
        return res;
    }

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