LeetCode 1101. The Earliest Moment When Everyone Become Friends

原题链接在这里：https://leetcode.com/problems/the-earliest-moment-when-everyone-become-friends/

题目：

In a social group, there are N people, with unique integer ids from 0 to N-1.

We have a list of logs, where each logs[i] = [timestamp, id_A, id_B] contains a non-negative integer timestamp, and the ids of two different people.

Each log represents the time in which two different people became friends.  Friendship is symmetric: if A is friends with B, then B is friends with A.

Let's say that person A is acquainted with 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. Return -1 if there is no such earliest time.

 

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 friend anything happens.
The sixth event occurs at timestamp = 20190301 and after 0 and 3 become friends we have that all become friends.

Note:

1. 1 <= N <= 100
2. 1 <= logs.length <= 10^4
3. 0 <= logs[i][0] <= 10^9
4. 0 <= logs[i][1], logs[i][2] <= N - 1
5. It's guaranteed that all timestamps in logs[i][0] are different.
6. Logs are not necessarily ordered by some criteria.
7. logs[i][1] != logs[i][2]

题解：

If log[1] and log[2] do NOT have same ancestor, put them into same union. 

When count of unions become 1, that is the first time all people become friends and output time.

Time Complexity: O(mlogN). m = logs.length. find takes O(logN). With path compression and union by weight, amatorize O(1).

Space: O(N).

AC Java:

class Solution {
    public int earliestAcq(int[][] logs, int N) {
        UF uf= new UF(N);
        Arrays.sort(logs, (a, b)-> a[0]-b[0]);
        
        for(int [] log : logs){
            if(uf.find(log[1]) != uf.find(log[2])){
                uf.union(log[1], log[2]);   
            }
            
            if(uf.count == 1){
                return log[0];
            }
        }
        
        return -1;
    }
}

class UF{
    int [] parent;
    int [] size;
    int count;
    
    public UF(int n){
        this.parent = new int[n];
        this.size = new int[n];
        for(int i = 0; i<n; i++){
            parent[i] = i;
            size[i] = 1;
        }
        
        this.count = n;
    }
    
    public int find(int i){
        while(i != parent[i]){
            parent[i] = parent[parent[i]];
            i = parent[i];
        }
        
        return parent[i];
    }
    
    public void union(int p, int q){
        int i = find(p);
        int j = find(q);
        if(size[i] > size[j]){
            parent[j] = i;
            size[i] += size[j];
        }else{
            parent[i] = j;
            size[j] += size[i];
        }
        
        this.count--;
    }
}

类似Friend Circles.