clueless coding // TODO: be smarter

LeetCode 454. 4Sum II: C++ Solution



Problem Statement


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.

Function Signature (C++):
  int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D)

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

Outputs:
  2

We can create two different tuples:
  (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
  (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0



TL;DR Code Solution


int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
    unordered_map<int, int> sums;
    for (int i = 0; i < A.size(); i++) {
        for (int j = 0; j < B.size(); j++) {
            sums[A[i] + B[j]]++;
        }
    }

    int count = 0;
    for (int i = 0; i < C.size(); i++) {
        for (int j = 0; j < D.size(); j++) {
            if (sums.find(-(C[i] + D[j])) != sums.end()) {
                count += sums[-(C[i] + D[j])];
            }
        }
    }

    return count;
}