// 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 -2^28 to 2^28 - 1 
 * and the result is guaranteed to be at most 2^31 - 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
 * 
 * Tags: Binary Search, Hash Table
 * 
 * Similar Problems: (M) 4Sum
 * 
 * Author: Kuang Qin
 */

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int n = A.size();
        
        // only count is needed, use a hash map for faster search
        unordered_map<int, int> map;
        
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                sum = -(A[i] + B[j]);
                map[sum]++; // create hash map using the -sum of A and B
            }
        }
        
        int res = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                sum = C[i] + D[j];
                if (map.find(sum) != map.end()) // look up in hash map using the sum of C and D
                    res += map[sum];
            }
        }
        
        return res;
    }
};

int main()
{
    vector<int> A = {1, 2};
    vector<int> B = {-2, -1};
    vector<int> C = {-1, 2};
    vector<int> D = {0, 2};
    Solution sol;
    int res = sol.fourSumCount(A, B, C, D);
    cout << res << endl;
    cin.get();
    
    return 0;
}
Add Solution for 454 4Sum II
Add Solution for 454 4Sum II
