1101

1101. The Earliest Moment When Everyone Become Friends (Medium)

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

 

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.

Companies:
Google

Related Topics:
Array, Union Find

Similar Questions:

• Number of Provinces (Medium)

Solution 1. Union Find

// OJ: https://leetcode.com/problems/the-earliest-moment-when-everyone-become-friends/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class UnionFind {
    vector<int> id;
    int cnt;
public:
    UnionFind(int n) : id(n), cnt(n) {
        iota(begin(id), end(id), 0);
    }
    int find(int a) {
        return id[a] == a ? a : (id[a] = find(id[a]));
    }
    void connect(int a, int b) {
        int x = find(a), y = find(b);
        if (x == y) return;
        id[x] = y;
        --cnt;
    }
    int getCount() { return cnt; }
};
class Solution {
public:
    int earliestAcq(vector<vector<int>>& A, int n) {
        sort(begin(A), end(A));
        UnionFind uf(n);
        for (auto &log : A) {
            int t = log[0], u = log[1], v = log[2];
            uf.connect(u, v);
            if (uf.getCount() == 1) return t;
        }
        return -1;
    }
};