<p>You are given an integer <code>n</code> indicating there are <code>n</code> people numbered from <code>0</code> to <code>n - 1</code>. You are also given a <strong>0-indexed</strong> 2D integer array <code>meetings</code> where <code>meetings[i] = [x<sub>i</sub>, y<sub>i</sub>, time<sub>i</sub>]</code> indicates that person <code>x<sub>i</sub></code> and person <code>y<sub>i</sub></code> have a meeting at <code>time<sub>i</sub></code>. A person may attend <strong>multiple meetings</strong> at the same time. Finally, you are given an integer <code>firstPerson</code>.</p>

<p>Person <code>0</code> has a <strong>secret</strong> and initially shares the secret with a person <code>firstPerson</code> at time <code>0</code>. This secret is then shared every time a meeting takes place with a person that has the secret. More formally, for every meeting, if a person <code>x<sub>i</sub></code> has the secret at <code>time<sub>i</sub></code>, then they will share the secret with person <code>y<sub>i</sub></code>, and vice versa.</p>

<p>The secrets are shared <strong>instantaneously</strong>. That is, a person may receive the secret and share it with people in other meetings within the same time frame.</p>

<p>Return <em>a list of all the people that have the secret after all the meetings have taken place. </em>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1
<strong>Output:</strong> [0,1,2,3,5]
<strong>Explanation:
</strong>At time 0, person 0 shares the secret with person 1.
At time 5, person 1 shares the secret with person 2.
At time 8, person 2 shares the secret with person 3.
At time 10, person 1 shares the secret with person 5.​​​​
Thus, people 0, 1, 2, 3, and 5 know the secret after all the meetings.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3
<strong>Output:</strong> [0,1,3]
<strong>Explanation:</strong>
At time 0, person 0 shares the secret with person 3.
At time 2, neither person 1 nor person 2 know the secret.
At time 3, person 3 shares the secret with person 0 and person 1.
Thus, people 0, 1, and 3 know the secret after all the meetings.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1
<strong>Output:</strong> [0,1,2,3,4]
<strong>Explanation:</strong>
At time 0, person 0 shares the secret with person 1.
At time 1, person 1 shares the secret with person 2, and person 2 shares the secret with person 3.
Note that person 2 can share the secret at the same time as receiving it.
At time 2, person 3 shares the secret with person 4.
Thus, people 0, 1, 2, 3, and 4 know the secret after all the meetings.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= meetings.length &lt;= 10<sup>5</sup></code></li>
	<li><code>meetings[i].length == 3</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i </sub>&lt;= n - 1</code></li>
	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
	<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= firstPerson &lt;= n - 1</code></li>
</ul>

from collections import defaultdict, deque
class UnionFind:
    def __init__(self, n):
        self.parents = list(range(n))
        self.ranks = [0] * n
    
    def union(self, a, b):
        root_a = self.find(a)
        root_b = self.find(b)
        # print(a, b, root_a, root_b)
        if root_a == root_b:
            return
        if self.ranks[root_a] > self.ranks[root_b]:
            self.parents[root_b] = root_a
        elif self.ranks[root_a] < self.ranks[root_b]:
            self.parents[root_a] = root_b
        else:
            self.parents[root_b] = root_a
            self.ranks[root_a] += 1

    def find(self, a):
        if self.parents[a] == a:
            return a
        root_a = self.find(self.parents[a])
        self.parents[a] = root_a
        return root_a

class Solution:
    def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
        groups = defaultdict(set)
        meetings_for_people = [{} for _ in range(n)]
        
        for a, b, t in meetings:
            groups[t].add(a)
            groups[t].add(b)
            if t not in meetings_for_people[a]:
                meetings_for_people[a][t] = [(a, b, t)]
            else:
                meetings_for_people[a][t].append((a, b, t))
            if t not in meetings_for_people[b]:
                meetings_for_people[b][t] = [(a, b, t)]
            else:
                meetings_for_people[b][t].append((a, b, t))
            
        meeting_times = list(groups.keys())
        meeting_times.sort()

        finder = UnionFind(n)
        finder.union(0, firstPerson)
        root_0 = finder.find(0)

        for t in meeting_times:
            init_persons = deque([])
            visited = set(init_persons)
            for a in groups[t]:
                if finder.find(a) == root_0:
                    init_persons.append(a)
            while init_persons:
                a = init_persons.popleft()
                for meeting in meetings_for_people[a][t]:
                    x, y, _ = meeting
                    if x in visited and y in visited:
                        continue
                    finder.union(x, y)
                    if x == a:
                        visited.add(y)
                        init_persons.append(y)
                    else:
                        visited.add(x)
                        init_persons.append(x)
            
        res = []
        root_0 = finder.find(0)
        for i in range(n):
            if finder.find(i) == root_0:
                res.append(i)
        
        # print(finder.parents)

        return res

1991