andre - 1 year ago 74

C++ Question

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

I just came up with a solution that concatenates all the vectors and find the 4 sum. But I know there is a better solution. Would someone explain a better solution ? I just see codes using O(N^2) but I can't understand it.

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

This was my `O(n^2)`

solution:

The core idea is - if `W + X + Y + Z = 0`

then `W + X = -(Y + Z)`

.

```
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int n = A.size();
int result = 0;
unordered_map<int,int> sumMap1;
unordered_map<int,int> sumMap2;
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n; ++j) {
int sum1 = A[i] + B[j];
int sum2 = C[i] + D[j];
sumMap1[sum1]++;
sumMap2[sum2]++;
}
}
for(auto num1 : sumMap1) {
int number = num1.first;
if(sumMap2.find(-1 * number) != sumMap2.end()) {
result += num1.second * sumMap2[-1 * number];
}
}
return result;
}
```

I am going to add some explanation.

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**