454. 4Sum II

• Difficulty: Medium

• Topics: Hash Table, Binary Search

• Similar Questions:

  • 4Sum

Problem:

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:

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int, int> valueFreq;

        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < B.size(); ++j) {
                ++valueFreq[A[i] + B[j]];
            }
        }

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

        return count;
    }
};