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# 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 + nums2 + nums3 + nums4 = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1 + nums2 + nums3 + nums4 = 2 + (-1) + (-1) + 0 = 0``````

Example 2:

``````Input: nums1 = , nums2 = , nums3 = , nums4 = 
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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