<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:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        sort(begin(meetings), end(meetings), [](const auto& a, const auto& b) { return a[2] < b[2]; });
        unordered_set<int> result = {0, firstPerson};
        unordered_map<int, vector<int>> adj;
        for (int i = 0; i < size(meetings); ++i) {
            int x = meetings[i][0], y = meetings[i][1];
            adj[x].emplace_back(y);
            adj[y].emplace_back(x);
            if (i + 1 != size(meetings) && meetings[i + 1][2] == meetings[i][2]) {
                continue;
            }
            vector<int> q;
            for (const auto& [k, _] : adj) {
                if (result.count(k)) {
                    q.emplace_back(k);
                }
            }
            while (!empty(q)) {
                vector<int> new_q;
                for (const auto& u : q) {
                    for (const auto& v : adj[u]) {
                        if (result.count(v)) {
                            continue;
                        }
                        result.emplace(v);
                        new_q.emplace_back(v);
                    }
                }
                q = move(new_q);
            }
            adj.clear();
        }
        return {cbegin(result), cend(result)};
    }
};

// Time:  O(nlogn)
// Space: O(n)
class Solution2 {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        sort(begin(meetings), end(meetings), [](const auto& a, const auto& b) { return a[2] < b[2]; });
        unordered_set<int> result = {0, firstPerson};
        unordered_map<int, vector<int>> adj;
        for (int i = 0; i < size(meetings); ++i) {
            int x = meetings[i][0], y = meetings[i][1];
            adj[x].emplace_back(y);
            adj[y].emplace_back(x);
            if (i + 1 != size(meetings) && meetings[i + 1][2] == meetings[i][2]) {
                continue;
            }
            vector<int> stk;
            for (const auto& [k, _] : adj) {
                if (result.count(k)) {
                    stk.emplace_back(k);
                }
            }
            while (!empty(stk)) {
                int u = stk.back(); stk.pop_back();
                for (const auto& v : adj[u]) {
                    if (result.count(v)) {
                        continue;
                    }
                    result.emplace(v);
                    stk.emplace_back(v);
                }
            }
            adj.clear();
        }
        return {cbegin(result), cend(result)};
    }
};

// Time:  O(nlogn)
// Space: O(n)
class Solution3 {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        sort(begin(meetings), end(meetings), [](const auto& a, const auto& b) { return a[2] < b[2]; });
        UnionFind uf(n);
        uf.union_set(0, firstPerson);
        unordered_set<int> group;
        for (int i = 0; i < size(meetings); ++i) {
            int x = meetings[i][0], y = meetings[i][1];
            group.emplace(x);
            group.emplace(y);
            uf.union_set(x, y);
            if (i + 1 != size(meetings) && meetings[i + 1][2] == meetings[i][2]) {
                continue;
            }
            for (const auto& x: group) {
                if (uf.find_set(x) != uf.find_set(0)) {
                    uf.reset(x);
                }
            }
            group.clear();
        }
        vector<int> result;
        for (int i = 0; i < n; ++i) {
            if (uf.find_set(i) == uf.find_set(0)) {
                result.emplace_back(i);
            }
        }
        return result;
    }

private:
    class UnionFind {
    public:
        UnionFind(int n)
         : set_(n)
         , rank_(n)  {
            iota(set_.begin(), set_.end(), 0);
        }

        int find_set(int x) {
           if (set_[x] != x) {
               set_[x] = find_set(set_[x]);  // Path compression.
           }
           return set_[x];
        }

        bool union_set(int x, int y) {
            x = find_set(x), y = find_set(y);
            if (x == y) {
                return false;
            }
            if (rank_[x] > rank_[y]) {
                swap(x, y);
            }
            set_[x] = y;  // Union by rank.
            if (rank_[x] == rank_[y]) {
                ++rank_[y];
            }
            return true;
        }

        void reset(int x) {
            set_[x] = x;
            rank_[x] = 0;
         }

    private:
        vector<int> set_;
        vector<int> rank_;
    };
};
221251