<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:
    int find(vector<int>& dsu, int index) {
        vector<int> path;

        while (dsu[index] >= 0) {
            path.push_back(index);
            index = dsu[index];
        }

        for (int i : path) {
            dsu[i] = index;
        }

        return index;
    }

    void dsunion(vector<int>& dsu, int a, int b) {
        int ina = find(dsu, a);
        int inb = find(dsu, b);

        if (ina == inb) {
            return;
        }

        if (-dsu[ina] < -dsu[inb]) {
            swap(ina, inb);
        }

        dsu[ina] += dsu[inb];
        dsu[inb] = ina;
    }

    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        vector<int> dsu(n, -1);
        dsunion(dsu, 0, firstPerson);

        auto comp = [](vector<int>& a, vector<int>& b) {
            return a[2] < b[2]; 
        };

        sort(meetings.begin(), meetings.end(), comp);

        int low = 0;
        int high = 0;

        while (high < meetings.size()) {
            unordered_set<int> considered;
            while (high < meetings.size() && meetings[high][2] == meetings[low][2]) {
                dsunion(dsu, meetings[high][0], meetings[high][1]);
                considered.insert(meetings[high][0]);
                considered.insert(meetings[high][1]);

                high++;
            }

            for (int i : considered) {
                if (find(dsu, i) != find(dsu, 0)) {
                    dsu[i] = -1;
                }
            }

            low = high;
        }

        vector<int> out;

        for (int i = 0; i < dsu.size(); i++) {
            if (find(dsu, i) == find(dsu, 0)) {
                out.push_back(i);
            }
        }

        return out;
    }
};
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