<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 {
public:
class DisjointSet{
  unordered_map<int,int>rank,parent,size;
public:
  DisjointSet(vector<int>a){
    for(auto i:a){
        parent[i]=i;
        size[i]=1;
    }
   
  }
  int findUpar(int node){
    if(node==parent[node]) return node;
    return parent[node]=findUpar(parent[node]);
  }

  

    void unionbysize(int u,int v){
          int ulp_u=findUpar(u);
    int ulp_v=findUpar(v);
    if(ulp_u==ulp_v) return;
    if(size[ulp_u]<size[ulp_v]){
      parent[ulp_u]=ulp_v;
      size[ulp_v]+=size[ulp_u];
    }
    else if(size[ulp_u]>=size[ulp_v]){
      parent[ulp_v]=ulp_u;
      size[ulp_u]+=size[ulp_v];
    }

  
    }
      void reset(int node){
        size[node]=1;
        parent[node]=node;
    }
};

    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        // goes with time
        map<int,vector<vector<int>>>mp;
        map<int,int>mintime;
        vector<int>a;
        a.push_back(0);
        for(auto i:meetings){
            mp[i[2]].push_back({i[0],i[1]});
           
          
            
        }
        for(int i=0;i<n;i++){
            a.push_back(i);
        }
        int start=mintime[firstPerson];
        map<int,int>done;
        done[0]=1;
        done[firstPerson]=1;
        
        DisjointSet ds(a);
        ds.unionbysize(0,firstPerson);
        for(auto i:mp){
           // if(i.first<start) continue;
            vector<int>p;
            bool ok=false;
            for(auto j:i.second){
             
                if(done[j[0]] || done[j[1]]) ok=true;
            }
            //cout<<ok<<endl;
            if(!ok) continue;
            for(auto j:i.second) {ds.unionbysize(j[0],j[1]); 
              
            }

            for(auto j:i.second){
                  if(ds.findUpar(j[0])==ds.findUpar(0)) done[j[0]]=1;
                  else ds.reset(j[0]);
                if(ds.findUpar(j[1])==ds.findUpar(0)) done[j[1]]=1;
                    else ds.reset(j[1]);
            }
        }

        vector<int>ans;
        map<int,int>vis;
        int x=ds.findUpar(0);
        for(auto i:a){
            if(vis[i] ) continue;
            vis[i]=1;
         //   cout<<ds.findUpar(i)<<endl;
            if(ds.findUpar(i)==x) ans.push_back(i);
        }
        return ans;
    }
};
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