Find All People With Secret
<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>
Depth-First Search; Breadth-First Search; Union Find; Graph; Sorting
function findAllPeople(n: number, meetings: number[][], firstPerson: number): number[] {
    let parent: Array<number> = Array.from({ length: n + 1 }, (v, i) => i);
    parent[firstPerson] = 0;

    function findParent(index: number): number {
        if (parent[index] != index) parent[index] = findParent(parent[index]);
        return parent[index];
    }

    let map = new Map<number, Array<Array<number>>>();
    for (let meeting of meetings) {
        const time = meeting[2];
        let members: Array<Array<number>> = map.get(time) || new Array();
        members.push(meeting);
        map.set(time, members);
    }
    const times = [...map.keys()].sort((a, b) => a - b);
    for (let time of times) {
        // round 1
        for (let meeting of map.get(time)) {
            let [a, b] = meeting;
            if (!parent[findParent(a)] || !parent[findParent(b)]) {
                parent[findParent(a)] = 0;
                parent[findParent(b)] = 0;
            }
            parent[findParent(a)] = parent[findParent(b)];
        }
        // round 2
        for (let meeting of map.get(time)) {
            let [a, b] = meeting;
            if (!parent[findParent(a)] || !parent[findParent(b)]) {
                parent[findParent(a)] = 0;
                parent[findParent(b)] = 0;
            } else {
                parent[a] = a;
                parent[b] = b;
            }
        }
    }

    let ans = new Array<number>();
    for (let i = 0; i <= n; i++) {
        if (!parent[findParent(i)]) {
            ans.push(i);
        }
    }
    return ans;
}
