1101. The Earliest Moment When Everyone Become Friends

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], , , , .
The second event occurs at timestamp = 20190104 and after 3 and 4 become friends we have the following friendship groups [0,1], , [3,4], .
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], .
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. `2 <= N <= 100`
2. `1 <= logs.length <= 10^4`
3. `0 <= logs[i] <= 10^9`
4. `0 <= logs[i], logs[i] <= N - 1`
5. It's guaranteed that all timestamps in `logs[i]` are different.
6. `logs `are not necessarily ordered by some criteria.
7. `logs[i] != logs[i]`

1101. The Earliest Moment When Everyone Become Friends
``````#![allow(clippy::unreadable_literal)]
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;

struct UnionFind {
parent: Vec<usize>,
group: usize,
n: usize,
}

impl UnionFind {
fn new(n: usize) -> Self {
let parent = (0..n).collect();
UnionFind {
parent,
group: n,
n,
}
}

fn find(&mut self, i: usize) -> usize {
let j = self.parent[i];
if i == j {
i
} else {
let k = self.find(j);
self.parent[i] = k;
k
}
}

fn union(&mut self, mut i: usize, mut j: usize) -> usize {
i = self.find(i);
j = self.find(j);
if i != j {
self.parent[j] = i;
self.group -= 1;
}
self.group
}
}

type Log = (Reverse<i32>, usize, usize);

impl Solution {
fn earliest_acq(logs: Vec<Vec<i32>>, n: i32) -> i32 {
let mut queue: BinaryHeap<Log> = logs
.iter()
.map(|v| (Reverse(v), v as usize, v as usize))
.collect();
let n = n as usize;
let mut uf = UnionFind::new(n);
while let Some(log) = queue.pop() {
if uf.union(log.1, log.2) == 1 {
return (log.0).0;
}
}
-1
}
}

#[test]
fn test() {
let logs = vec_vec_i32![
[20190101, 0, 1],
[20190104, 3, 4],
[20190107, 2, 3],
[20190211, 1, 5],
[20190224, 2, 4],
[20190301, 0, 3],
[20190312, 1, 2],
[20190322, 4, 5]
];
let n = 6;
let res = 20190301;
assert_eq!(Solution::earliest_acq(logs, n), res);
}
``````