454. 4Sum II

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

 

Approach #1: burp force[Limted Time Exceeded]

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int len = A.size();
        int count = 0;
        for (int i = 0; i < len; ++i) {
            for (int j = 0; j < len; ++j) {
                for (int k = 0; k < len; ++k) {
                    for (int h = 0; h < len; ++h) {
                        if (A[i] + B[j] + C[k] + D[h] == 0)
                            count++;
                    }
                }
            }
        }
        return count;
    }
};

　　

Approach #2: Hash Table

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int count = 0;
        map<int, int> ab_sum;
        for (int a : A) {
            for (int b : B) {
                ++ab_sum[a+b];
            }
        }
        for (int c : C) {
            for (int d : D) {
                auto it = ab_sum.find(0-c-d);
                if (it != ab_sum.end()) count += it->second;
            }
        }
        return count;
    }
};

Runtime: 248 ms, faster than 25.52% of C++ online submissions for 4Sum II.