<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 
{
private:
    vector<bool> isKnown;
    vector<int> vis;

public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) 
    {
        //cout<<n<<" "<<meetings.size()<<endl;
        
        map<int, vector<vector<int>>> groups;
        for (auto& it : meetings)
        {
            int a = it[0];
            int b = it[1];
            int t = it[2];
            
            groups[t].push_back({ a,b });
        }

        isKnown.resize(n, false);
        isKnown[0] = true;
        isKnown[firstPerson] = true;

        vis.resize(n, false);

        for (auto& [time, it] : groups)
        {
            unordered_map<int, unordered_set<int>> g;
            //vector<unordered_set<int>> g;
            unordered_set<int> initial;
            build(it, g, initial);
            
            if(g.empty() || initial.empty()) continue;
            
            //int cnt = 0;
            // for(int i = 0; i < n; ++i)
            // {
            //     cnt += g[i].size();
            //     //cout<<i<<" : ";
            //     //for(auto& j : g[i]) cout<<j<<" "; cout<<endl;
            // }
            //cout << time << " " << g.size() << " " << initial.size() <<" | " << cnt <<endl;
            
            bfs(g, initial);
        }

        vector<int> res;
        for (int i = 0; i < n; ++i)
        {
            if (isKnown[i]) res.push_back(i);
        }
        
        return res;
    }

    void build(vector<vector<int>>& edges, unordered_map<int, unordered_set<int>>& g, unordered_set<int>& initial)
    {
        bool hasKnown = false;
        //int maxNode = 0;
        for (auto& e : edges)
        {
            int a = e[0];
            int b = e[1];
            //maxNode = max(maxNode, max(a, b));
            
            if (isKnown[a] || isKnown[b]) hasKnown = true; 
        }
        
        if(!hasKnown) return;
        
        //g.resize(maxNode + 1);
        for (auto& e : edges)
        {         
            int a = e[0];
            int b = e[1];
            if(isKnown[a] && isKnown[b]) continue;
            
            g[a].insert(b);
            g[b].insert(a);
            
            if (isKnown[a]) initial.insert(a);
            if (isKnown[b]) initial.insert(b);
        }
    }
    
    void bfs(unordered_map<int, unordered_set<int>>& g, unordered_set<int> initial)
    {
        if(initial.empty() || g.empty()) return;
        
        fill(begin(vis), end(vis), false);
        
        queue<int> q;
        for(int i : initial) 
        {            
            q.push(i);
            vis[i] = true;
        }
        
        while(!q.empty())
        {
            int cur = q.front();
            q.pop();
            //cout<<cur<<endl;
            
            isKnown[cur] = true;
            for(int next : g[cur])
            {
                if(vis[next]) continue;
                vis[next] = true;
                
                q.push(next);
            }
        }                
    }
    
    void dfs(vector<unordered_set<int>>& g, int cur)
    {
        //cout<<cur<<endl;
        vis[cur] = true;
        
        isKnown[cur] = true;
        for (auto& next : g[cur])
        {
            if (vis[next]) continue;
            //vis[next] = true;
            
            dfs(g, next);
        }
    }
};
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