<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>

class Solution {

    void findMinTimeToDiscoverSecret(
        unordered_map< int, list< pair<int, int> > >& graph,
        vector<int>& discoveryTime,
        int firstPerson
    ){
        // Declare the required data structure
        multimap< int, int> pq;
        unordered_map<int, int> visited;

        // Initialise starting conditions
        discoveryTime[0] = 0;
        discoveryTime[firstPerson] = 0;
        visited[0] = 1;
        visited[firstPerson] = 1;   
        pq.insert(make_pair(0, 0));     
        pq.insert(make_pair(0, firstPerson));  

        // Perform BFS
        while(pq.size() != 0){
            // Identify the next node discovered
            int currentDiscoveryTime = pq.begin()->first;
            int current = pq.begin()->second;
            pq.erase(pq.begin());
            if(visited[current] == 2)continue;
            visited[current] = 2;
            discoveryTime[current] = currentDiscoveryTime;

            // Relax its neighbors 
            for(auto neighborDetails: graph[current]){
                int neighbor = neighborDetails.first;
                int meetingTime = neighborDetails.second;
                if(meetingTime < currentDiscoveryTime)continue;
                if(visited[neighbor] == 2)continue;
                if(discoveryTime[neighbor] <= meetingTime)continue;
                discoveryTime[neighbor] = meetingTime;
                visited[neighbor] = 1; 
                pq.insert(make_pair(discoveryTime[neighbor], neighbor));             
            }

        }   

    }


public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        // Create required Graph Data Structure
        unordered_map< int, list< pair<int, int> > > graph; 
        for(auto meeting: meetings){
            graph[meeting[0]].push_back(make_pair(meeting[1], meeting[2]));
            graph[meeting[1]].push_back(make_pair(meeting[0], meeting[2]));
        }

        // find min time to discover secret 
        vector<int> discoveryTime(n, INT_MAX);
        findMinTimeToDiscoverSecret(graph, discoveryTime, firstPerson);
        
        // find people with secret
        vector<int> result(0);
        for(int counter = 0; counter < n; counter++)
           if(discoveryTime[counter] != INT_MAX)result.push_back(counter);
        return result;
        
    }
};
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