4Sum II

 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

class Solution {
public:
 int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) 
    {
        
        unordered_map<int,int> mp;
        int n=A.size();
        int ans=0,sum;
        for(int i=0;i<n;++i)
        {
            for(int j=0;j<n;++j)
            {  
                sum=A[i]+B[j];
                mp[sum]++;
            }
        }
        
    
        for(int i=0;i<n;++i)
        {
            for(int j=0;j<n;++j)
            {
                 sum=-(C[i]+D[j]);
                if(mp.find(sum)!=mp.end())
                    ans+=mp[sum];
            }
        }
        return ans;
        
    }
};

hERE IS JAVA SOLUTION
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    Map<Integer, Integer> map = new HashMap<>();
    
    for(int i=0; i<C.length; i++) {
        for(int j=0; j<D.length; j++) {
            int sum = C[i] + D[j];
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
    }
    
    int res=0;
    for(int i=0; i<A.length; i++) {
        for(int j=0; j<B.length; j++) {
            res += map.getOrDefault(-1 * (A[i]+B[j]), 0);
        }
    }
    
    return res;
}

Time complexity:  O(n^2)
Space complexity: O(n^2)

HERE IS PAYTHON SOLUTION:
def fourSumCount(self, A, B, C, D):
    AB = collections.Counter(a+b for a in A for b in B)
    return sum(AB[-c-d] for c in C for d in D)