<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>

#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        // Map to store meetings by their time
        map<int, vector<pair<int, int>>> meetingMap;
        for (const auto& meeting : meetings) {
            int time = meeting[2];
            int person1 = meeting[0];
            int person2 = meeting[1];
            meetingMap[time].emplace_back(person1, person2);
        }

        vector<bool> knows(n, false);
        knows[0] = true; // Person 0 knows the secret
        knows[firstPerson] = true; // First person knows the secret
        
        // Process meetings by time
        auto it = meetingMap.begin();
        while (it != meetingMap.end()) {
            unordered_map<int, vector<int>> graph;
            unordered_set<int> peopleWithSecret;
            vector<int> nodesInCurrentTime;
            
            // Build the graph for the current timestamp
            for (const auto& meeting : it->second) {
                int p1 = meeting.first;
                int p2 = meeting.second;
                if (knows[p1] || knows[p2]) {
                    peopleWithSecret.insert(p1);
                    peopleWithSecret.insert(p2);
                }
                graph[p1].push_back(p2);
                graph[p2].push_back(p1);
            }

            // Perform BFS to propagate the secret
            unordered_set<int> visited;
            for (int person : peopleWithSecret) {
                if (visited.find(person) == visited.end()) {
                    queue<int> q;
                    q.push(person);
                    visited.insert(person);
                    knows[person] = true;
                    
                    while (!q.empty()) {
                        int current = q.front();
                        q.pop();
                        for (int neighbor : graph[current]) {
                            if (visited.find(neighbor) == visited.end()) {
                                visited.insert(neighbor);
                                knows[neighbor] = true;
                                q.push(neighbor);
                            }
                        }
                    }
                }
            }

            ++it;
        }
        
        // Collect all nodes with the secret
        vector<int> result;
        for (int i = 0; i < n; ++i) {
            if (knows[i]) {
                result.push_back(i);
            }
        }

        // Sort the result
        sort(result.begin(), result.end());
        return result;
    }
};

352019