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.

Approach:

• Take the arrays A and B, and compute all the possible sums of two elements.

• Put the sum in the Hash map, and increase the hash map value if more than 1 pair sums to the same value.

• Compute all the possible sums of the arrays C and D. If the hash map contains the opposite value of the current sum, increase the count of four elements sum to 0 by the counter in the map.

class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        int n = A.length;
        int result = 0;
        Map<Integer, Integer> map = new HashMap<>();
        
        for(int a : A){
            for(int b : B){
                int sumAB = a + b;
                map.put(sumAB, map.getOrDefault(sumAB, 0) + 1);
            }
        }
        for(int c : C){
            for(int d : D){
                int sumCD = -(c + d);
                if(map.containsKey(sumCD)){
                    result += map.get(sumCD);
                }
            }
        }
        return result;
    }
}

Time: O(n2) Space: O(n)