4Sum II

Description

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

Code

Time Complexity: $O(N^2)$, Space Complexity: $O(N^2)$

Brute force 會是 $O(N^4)$，但是可以用 Two Sum 的概念配合 hash table 降低至 $O(N^2)$。

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        unordered_map<int, int> mp;
        for(auto i: nums1) {
            for(auto j: nums2) {
                mp[i + j]++;
            }
        }
 
        int res = 0;
        for(auto i: nums3) {
            for(auto j: nums4) {
                res += mp[-(i + j)];
            }
        }
 
        return res;
    }
};

Source

• 4Sum II - LeetCode