# leetcode\_454

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A\[i] + B\[j] + C\[k] + D\[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input: A = \[ 1, 2] B = \[-2,-1] C = \[-1, 2] D = \[ 0, 2]

Output: 2

Explanation: The two tuples are: 1. (0, 0, 0, 1) -> A\[0] + B\[0] + C\[0] + D\[1] = 1 + (-2) + (-1) + 2 = 0 2. (1, 1, 0, 0) -> A\[1] + B\[1] + C\[0] + D\[0] = 2 + (-1) + (-1) + 0 = 0

## Solutions

1. **hashmap O(n2)**

```cpp
class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int, int> sum;
        for (int i = 0; i < A.size(); i++)
            for (int j = 0; j < B.size(); j++) {
                sum[A[i] + B[j]]++;
            }

        int res = 0;
        for (int i = 0; i < C.size(); i++)
            for (int j = 0; j < D.size(); j++) {
                int s = C[i] + D[j];
                if (sum.count(-s))
                    res += sum[-s];
            }

        return res;
    }
};
```