<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 {
    public:
        vector<int> parent, size;

        DisjointSet(int n) {
            parent.resize(n);
            size.resize(n, 1);
            for (int i = 0; i < n; i++)
                parent[i] = i;
        }
        int findUPar(int x) {
            if (parent[x] == x)
                return x;
            else
                return parent[x] = findUPar(parent[x]);
        }
        void UnionBySize(int x, int y) {
            int p1 = findUPar(x);
            int p2 = findUPar(y);
            if (p1 == p2)
                return;
            if (size[p1] < size[p2]) {
                parent[p1] = p2;
                size[p2] = size[p2] + size[p1];
            } else {
                parent[p2] = p1;
                size[p1] = size[p1] + size[p2];
            }
        }
    };
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings,
                              int firstPerson) {
        int maxi = 0;
        for (auto it : meetings)
            maxi = max(maxi, it[2]);
        vector<vector<pair<int, int>>> v(maxi + 1);
        vector<int> ans;
        unordered_set<int> s;
        for (auto it : meetings)
            v[it[2]].push_back({it[0], it[1]});
        DisjointSet ds(n);
        ds.UnionBySize(0, firstPerson);
        for (int i = 1; i <= maxi; i++) {
            s.clear();
            for (auto it : v[i]) {
                ds.UnionBySize(it.first, it.second);
                s.insert(it.first);
                s.insert(it.second);
            }
            for (auto it : s) {
                if (ds.findUPar(it) != ds.findUPar(0)) {
                    ds.parent[it] = it;
                    ds.size[it] = 1;
                }
            }
        }
        for (int i = 0; i < n; i++) {
            if (ds.findUPar(i) == ds.findUPar(0))
                ans.push_back(i);
        }
        return ans;
    }
};
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