<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

# class UnionFind:
#     def __init__(self, n):
#         self.parent = [i for i in range(n)]
#         self.knows = [False for i in range(n)]
#         self.rank = [1 for i in range(n)]

#     def set(self, x, value):
#         self.knows[x] = value

#     def know(self, x):
#         x_parent = self.find(x)

#         return self.knows[x_parent]

#     def get_true_indices(self, parent):
#         ans = [0]

#         for i in range(1, len(self.parent)):
#             if self.find(i) == self.find(parent) and self.knows[self.find(i)]:
#                 ans.append(i)

#         return ans

#     def find(self, x):
#         if self.parent[x] != x:
#             self.parent[x] = self.find(self.parent[x])

#         return self.parent[x]

#     def union(self, x, y):
#         y_parent = self.find(y)
#         x_parent = self.find(x)

#         if self.rank[y_parent] > self.rank[x_parent]:
#             self.parent[x_parent] = y_parent
#             self.knows[x_parent] = self.knows[x_parent] or self.knows[y_parent]
#         elif self.rank[x_parent] > self.rank[y_parent]:
#             self.parent[y_parent] = x_parent
#             self.knows[y_parent] = self.knows[x_parent] or self.knows[y_parent]

#         else:
#             self.parent[x_parent] = y_parent
#             self.rank[y_parent] += 1
#             self.knows[x_parent] = self.knows[x_parent] or self.knows[y_parent]

class Solution:
    def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
        meeting_time_to_meeting = defaultdict(list)
        # uf = UnionFind(n)

        # uf.set(0, True)
        # uf.set(firstPerson, True)

        know_array = [False for i in range(n)]
        know_array[0] = True
        know_array[firstPerson] = True

        for meeting in meetings:
            meeting_time_to_meeting[meeting[2]].append(meeting)

        times = list(meeting_time_to_meeting.keys())
        times.sort()

        for time in times:
            meetings_for_time = meeting_time_to_meeting[time]

            graph = defaultdict(set)
            knows = set()

            for x, y, t in meetings_for_time:
                if know_array[x]:
                    knows.add(x)
                elif know_array[y]:
                    knows.add(y)
                graph[x].add(y)
                graph[y].add(x)

            if len(knows) > 0:
                queue = list(knows)
                seen_nodes = set()
                # everyone becones true
                while queue:
                    node = queue.pop(0)

                    if node in seen_nodes:
                        continue

                    seen_nodes.add(node)

                    know_array[node] = True

                    for neighbor in graph[node]:
                        queue.append(neighbor)

                # for node, neighbors in graph.items():
                #     # If one person at the time knows them everyone one will know, so I think 
                #     # we need to have a way to see if at least 1 person in this group knows at this time
                #     for neighbor in neighbors:
                #         if uf.know(node) or uf.know(neighbor):
                #             uf.union(node, neighbor)


        # return uf.get_true_indices(firstPerson)
        ans = []

        for i in range(len(know_array)):
            if know_array[i]:
                ans.append(i)

        return ans
        
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