[LeetCode] 1101. The Earliest Moment When Everyone Become Friends

Posted 2023-01-25 CNoodle

There are n people in a social group labeled from 0 to n - 1. You are given an array logs where logs[i] = [timestampi, xi, yi] indicates that xi and yi will be friends at the time timestampi.

Friendship is symmetric. That means if a is friends with b, then b is friends with a. Also, person a is acquainted with a person b if a is friends with b, or a is a friend of someone acquainted with b.

Return the earliest time for which every person became acquainted with every other person. If there is no such earliest time, return -1.

Example 1:

Input: logs = [[20190101,0,1],[20190104,3,4],[20190107,2,3],[20190211,1,5],[20190224,2,4],[20190301,0,3],[20190312,1,2],[20190322,4,5]], n = 6
Output: 20190301
Explanation: 
The first event occurs at timestamp = 20190101, and after 0 and 1 become friends, we have the following friendship groups [0,1], [2], [3], [4], [5].
The second event occurs at timestamp = 20190104, and after 3 and 4 become friends, we have the following friendship groups [0,1], [2], [3,4], [5].
The third event occurs at timestamp = 20190107, and after 2 and 3 become friends, we have the following friendship groups [0,1], [2,3,4], [5].
The fourth event occurs at timestamp = 20190211, and after 1 and 5 become friends, we have the following friendship groups [0,1,5], [2,3,4].
The fifth event occurs at timestamp = 20190224, and as 2 and 4 are already friends, nothing happens.
The sixth event occurs at timestamp = 20190301, and after 0 and 3 become friends, we all become friends.

Example 2:

Input: logs = [[0,2,0],[1,0,1],[3,0,3],[4,1,2],[7,3,1]], n = 4
Output: 3
Explanation: At timestamp = 3, all the persons (i.e., 0, 1, 2, and 3) become friends.

Constraints:

• 2 <= n <= 100
• 1 <= logs.length <= 104
• logs[i].length == 3
• 0 <= timestampi <= 109
• 0 <= xi, yi <= n - 1
• xi != yi
• All the values timestampi are unique.
• All the pairs (xi, yi) occur at most one time in the input.

彼此熟识的最早时间。

题目的中文翻译参见这里。题目给的 logs 是一些人两两认识的关系和他们成为朋友的时间戳，要我们求的是最早什么时候所有人都互相认识。

思路是并查集。这道题类似547题，一开始每个人都自成一组，每个人都只认识自己。之后我们遍历 logs（注意 logs 需要按照时间戳排序），根据 logs 里给出的信息我们把人放到一些连通分量里。每处理完一条 log，我们就检查连通分量的个数。当连通分量 == 1 的时候，就说明所有人都被合并在一组里了，返回此时的时间戳。

时间O(nlogn) - n个点，每个点找自己的父节点的复杂度 ≈ logn

空间O(n)

Java实现

 class Solution 
     public int earliestAcq(int[][] logs, int n) 
         Arrays.sort(logs, (a, b) -> (a[0] - b[0]));
         UF uf = new UF(n);
         for (int[] log : logs) 
             int time = log[0];
             int a = log[1];
             int b = log[2];
             uf.union(a, b);
             if (uf.count == 1) 
                 return time;
             
         
         return -1;
     
 
 
 class UF 
     int[] parent;
     int count;
     
     public UF(int n) 
         parent = new int[n];
         for (int i = 0; i < n; i++) 
             parent[i] = i;
         
         count = n;
     
     
     public int find(int i) 
         if (i == parent[i]) 
             return i;
         
         return parent[i] = find(parent[i]);
     
     
     public void union(int p, int q) 
         int pRoot = find(p);
         int qRoot = find(q);
         if (pRoot != qRoot) 
             parent[pRoot] = qRoot;
             count--;
         
     
 

 

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