<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 disjoint_set
{
    int* size;
    int* parent;

    public:
    disjoint_set(int n)
    {//constructor, using plain arrays for s p e e d
        size = new int[n];
        for (int i = 0; i <= n-1; i++) size[i] = 1;

        parent = new int[n];
        for (int i = 0; i <= n-1; i++) parent[i] = i;
    }

    ~disjoint_set()
    {//destructor, memory cleanup
        delete[] size;
        delete[] parent;
    }

    int find_parent(int u)
    {//path compression
        if (u == parent[u]) return u;
        else return parent[u] = find_parent(parent[u]);
    }

    void union_size(int u, int v)
    {//union by size
        int uP = find_parent(u), vP = find_parent(v);
        if (uP == vP) return; //already union
        
        if (size[uP] >= size[vP])
        {//left bias dsu
            size[uP] += size[vP];
            parent[vP] = uP; 
        }
        else
        {
            size[vP] += size[uP];
            parent[uP] = vP;
        }
    }

    int set_size(int u)
    {//return num elements in a set
        return size[find_parent(u)];
    }

    void reset(int u)
    {//isolate element from its set
        parent[u] = u;
        size[u] = 1;
    }
};

class Solution {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson)
    {
        int maxTime = 0;
        for (vector<int> &meeting : meetings) maxTime = max(maxTime, meeting[2]);

        vector<vector<vector<int>>> peopleTime(maxTime + 1);    // edges at a partcular time
        for (vector<int> &meeting : meetings) peopleTime[meeting[2]].push_back({meeting[0], meeting[1]});

        disjoint_set secret(n);
        secret.union_size(0, firstPerson); //left bias in my dsu

        for (vector<vector<int>> &time : peopleTime)
        {
            if (time.empty()) continue;

            unordered_set<int> people;
            for (vector<int> &meeting : time)
            {
                people.insert(meeting[0]), people.insert(meeting[1]);
                secret.union_size(meeting[0], meeting[1]);
            }

            for (int p : people) if (secret.find_parent(0) != secret.find_parent(p)) secret.reset(p);
        }

        vector<int> ans;
        for (int i = 0; i <= n-1; i++) if (secret.set_size(i) > 1) ans.push_back(i);
        return ans;
    }    
};
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