<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>

bool compare(vector<int>&v1,vector<int>&v2){
   
    return v1[2]<v2[2];
}
class Solution {
public:
    int find(int i,vector<int>&parent){
        int a=parent[i];
        if(a==i){
            return i;
        }
        return parent[i]=find(a,parent);
    }
    void union2(int a,int b,vector<int>&parent,vector<int>&size){
        int i=find(a,parent);
        int j=find(b,parent);
        if(i==j){
            return;
        }
        if(i==0&&j!=0){
            size[i]+=size[j];
            parent[j]=i;
        }
        else{
            size[j]+=size[i];
            parent[i]=j;
        }
    }

    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int first) {
        sort(meetings.begin(),meetings.end(),compare);
        vector<int>ans;
        int m=meetings.size();
        vector<int>parent(n);
        vector<int>size(n,1);
        size[0]++;
        parent[0]=0;
        parent[first]=0;
        for(int i=1;i<n;i++){
            if(i!=first){
                parent[i]=i;
            }
        }
        int i=0;
        while(i<m){
            int j=i;
            vector<pair<int,int>>v;
            while(j<m&&meetings[j][2]==meetings[i][2]){
                int a=meetings[j][0];
                int b=meetings[j][1];
                j++;
                union2(a,b,parent,size);
                v.push_back({a,b});
            }
            i=j;
            for(auto p:v){
                int a=p.first;
                int b=p.second;
                if(find(a,parent)!=0){
                    parent[a]=a;
                }
                if(find(b,parent)!=0){
                    parent[b]=b;
                }
            }
        }
       
        for(int p=0;p<n;p++){
            if(find(p,parent)==0){
                ans.push_back(p);
            }
        }
        return ans;
    }
};
145544