4SumII.java

/*
 * Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:

0 <= i, j, k, l < n
nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0
 

Example 1:

Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
Output: 2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
Example 2:

Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1
 

Constraints:

n == nums1.length
n == nums2.length
n == nums3.length
n == nums4.length
1 <= n <= 200
-228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228
 */

 class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        Map<Integer, Integer> map = new HashMap();
        for(int d: nums4){
            for(int c: nums3){
                map.put(c+d, map.getOrDefault(c+d, 0)+1);
            }
        }
        int count = 0;
        for(int a: nums1){
            for(int b: nums2){
                count+= map.getOrDefault(-(a+b),0);
            }
        }
        return count;

    }
}





    // BASIC SOLUTION:
    //     int count = 0;
    //     for(int a: nums1){
    //         for(int b: nums2){
    //             for(int c: nums3){
    //                 for(int d: nums4){
    //                     if(a+b+c+d ==0) count++;

    //                 }
    //             }
    //         }
    //     }
    //     return count;
        
    // }

// Runtime 135 ms Beats 75.21%
// Memory 43.7 MB Beats 70.60%