4sum2.js

/* 4Sum II

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
*/

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @param {number[]} nums3
 * @param {number[]} nums4
 * @return {number}
 */
 var fourSumCount = function(nums1, nums2, nums3, nums4) {
    let tuples = 0;
    let n = nums1.length; 
    let tuplesSum = {};
    
    for(let i = 0; i < n; i++){
        for(let j = 0; j < n ; j++){
            let sum = nums1[i] + nums2[j];            
            if(sum in tuplesSum){
                tuplesSum[sum]++;
            }else{
                tuplesSum[sum] = 1;
            }
        }
    }
    
    for(let i = 0; i < n; i++){
        for(let j = 0; j < n ; j++){
            let sum = -1 * (nums3[i] + nums4[j]);
            if(tuplesSum[sum]){
                tuples += tuplesSum[sum];
            }
        }
    }
    
    return tuples;
};