## 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 = 6Output:20190301Explanation: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:**

`2 <= N <= 100`

`1 <= logs.length <= 10^4`

`0 <= logs[i][0] <= 10^9`

`0 <= logs[i][1], logs[i][2] <= N - 1`

- It's guaranteed that all timestamps in
`logs[i][0]`

are different. `logs`

are not necessarily ordered by some criteria.`logs[i][1] != logs[i][2]`

## Rust Solution

```
#![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[0]), v[1] as usize, v[2] 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);
}
```

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