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

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Idea : The 'getOrDefault' method reduces the use of one 'get' method

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Code : 

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer,Integer> AB=new HashMap<Integer, Integer>();
        int length = A.length;
        for(int i=0;i<length;i++){
            for(int j=0;j<length;j++){
                int temp1=A[i]+B[j];
                AB.put(temp1,AB.getOrDefault(temp1,0)+1);
            }
        }
        int res=0;
        for(int i=0;i<length;i++)
            for(int j=0;j<length;j++)
                res+=AB.getOrDefault(-(C[i]+D[j]),0);
        return res;
    }
}