LeetCode_ Hash table_ Medium_ 454. adding four numbers II

1. subject

Here are four integer arrays nums1、nums2、nums3 and nums4 , The length of the array is n , Please calculate how many tuples there are (i, j, k, l) To meet the ：
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
explain ：
Two tuples are as follows ：
(0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
(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

Tips ：
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

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/4sum-ii

2. Ideas

（1） The law of violent exhaustion
This method is simple and easy to think of , Use 4 individual for Loop enumerates all tuple combinations and makes judgments , Using variables res To record qualified tuples . But the time complexity of this method is too high （ by O(n⁴)）, Obviously this is LeetCode Will be given when submitting “ Time limit exceeded ” A hint of ！

（2） Hashtable _ grouping
Train of thought reference Official solution to this problem .

3. Code implementation （Java）

// Ideas 1———— The law of violent exhaustion 
class Solution {
    
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
    
        int n = nums1.length;
        int res = 0;
        for (int i = 0; i < n; i++) {
    
            for (int j = 0; j < n; j++) {
    
                for (int k = 0; k < n; k++) {
    
                    for (int l = 0; l < n; l++) {
    
                        if (nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0) {
    
                            res++;
                        }
                    }
                }
            }
        }
        return res;
    }
}

// Ideas 2———— Hashtable _ grouping 
class Solution {
    
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
    
        int res = 0;
        Map<Integer, Integer> cnt12 = new HashMap<>();
        for (int num1 : nums1) {
    
            for (int num2 : nums2) {
    
                cnt12.put(num1 + num2, cnt12.getOrDefault(num1 + num2, 0) + 1);
            }
        }
        for (int num3 : nums3) {
    
            for (int num4 : nums4) {
    
                if (cnt12.containsKey(-num3 - num4)) {
    
                    res += cnt12.get(-num3 - num4);
                }
            }
        }
        return res;
    }
}