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

Solution

C# Program

public class Solution {
    public int FourSumCount(int[] A, int[] B, int[] C, int[] D) {
        var map = new Dictionary<int,int>();
        int result = 0;
        for(int i=0;i<C.Length;i++)
        {
            for(int j=0;j<D.Length;j++)
            {
                int sum = C[i] + D[j];
                if(map.ContainsKey(sum))
                {
                    map[sum]++;
                }
                else{
                    map.Add(sum,1);
                }
            }
        }
        
        for(int i=0;i<A.Length;i++)
        {
            for(int j=0;j<B.Length;j++)
            {
                int sum = A[i] + B[j];
                if(map.ContainsKey(-1 * sum))
                {
                    result += map[-1 * sum];
                }
            }
        }
        
        return result;
    }
}

Time Complexity: O(m * n)

Space Complexity: O(m * n)

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