<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>

#pragma GCC optimize("O3", "unroll-loops")
class UnionFind {
    vector<int> root, rank;

public:
    UnionFind(int n) : root(n), rank(n) {
        rank.assign(n, 1);
        iota(root.begin(), root.end(), 0);
    }

    int Find(int x) {
        if (x == root[x])
            return x;
        else
            return root[x] = Find(root[x]);
    }

    void Union(int x, int y) {
        int rX = Find(x), rY = Find(y);
        if (rX == rY)
            return;
        if (rank[rX] > rank[rY])
            swap(rX, rY);
        root[rX] = rY;
        if (rank[rX] == rank[rY])
            rank[rY]++;
    }
    bool connected(int x, int y) { return Find(x) == Find(y); }

    void reset(int x) { // very useful for removing edges
        root[x] = x;
        rank[x] = 1;
    }
};

class Solution {
public:
    using int2 = pair<int, int>;
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings,
                              int firstPerson) {
        // 1 <= time_i <= 10^5, Counting sort is suitable
        vector<int2>
            meet_time[100001]; // meet_time[t] = {(x,y) has meeting at t}
        int tMax = -1;
        for (auto& meet : meetings) {
            int x = meet[0], y = meet[1], t = meet[2];
            meet_time[t].emplace_back(x, y); // 1 pair is enough
                                             // meet_time[t].emplace_back(y, x);
            tMax = max(tMax, t);
        }
        UnionFind uf(n);
        uf.Union(0, firstPerson); // add edge (0, firstPerson)
        for (int t = 0; t <= tMax; t++) {
            for (auto& [x, y] : meet_time[t])
                uf.Union(x, y);
            for (auto& [x, y] : meet_time[t]) { // Important part
                if (!uf.connected(0, x)) {
                    // If x is not connected to 0, remove the edge
                    uf.reset(x);
                    uf.reset(y);
                }
            }
        }
        vector<int> list = {0};
        for (int i = 1; i < n; i++)
            if (uf.connected(0, i))
                list.push_back(i);

        return list;
    }
};

auto init = []() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();
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