Problem: Find All People With Secret

You are given an integer `n` indicating there are `n` people numbered from `0`
to `n - 1`. You are also given a 0-indexed 2D integer array `meetings` where
`meetings[i] = [xi, yi, timei]` indicates that person `xi` and person `yi`
have a meeting at `timei`. A person may attend multiple meetings at the same
time. Finally, you are given an integer `firstPerson`.

Person `0` has a secret and initially shares the secret with a person
`firstPerson` at time `0`. 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 `xi` has the secret at `timei`, then they will share the
secret with person `yi`, and vice versa.

The secrets are shared instantaneously. That is, a person may receive the
secret and share it with people in other meetings within the same time frame.

Return a list of all the people that have the secret after all the meetings
have taken place. You may return the answer in any order.


Python solution:
import math
import itertools
import bisect
import collections
import string
import heapq
import functools
import sortedcontainers
from typing import List, Dict, Tuple, Iterator

class UnionFind:
  def __init__(self, n: int):
    self.id = list(range(n))
    self.rank = [0] * n

  def unionByRank(self, u: int, v: int) -> None:
    i = self._find(u)
    j = self._find(v)
    if i == j:
      return
    if self.rank[i] < self.rank[j]:
      self.id[i] = j
    elif self.rank[i] > self.rank[j]:
      self.id[j] = i
    else:
      self.id[i] = j
      self.rank[j] += 1

  def connected(self, u: int, v: int) -> bool:
    return self._find(self.id[u]) == self._find(self.id[v])

  def reset(self, u: int) -> None:
    self.id[u] = u

  def _find(self, u: int) -> int:
    if self.id[u] != u:
      self.id[u] = self._find(self.id[u])
    return self.id[u]


class Solution:
  def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
    uf = UnionFind(n)
    timeToPairs = collections.defaultdict(list)

    uf.unionByRank(0, firstPerson)

    for x, y, time in meetings:
      timeToPairs[time].append((x, y))

    for _, pairs in sorted(timeToPairs.items(), key=lambda x: x[0]):
      peopleUnioned = set()
      for x, y in pairs:
        uf.unionByRank(x, y)
        peopleUnioned.add(x)
        peopleUnioned.add(y)
      for person in peopleUnioned:
        if not uf.connected(person, 0):
          uf.reset(person)

    res=[]
    for i in range(n):
      if uf.connected(i, 0):
        res.append(i)
    return res


Java solution:
class UnionFind {
  public UnionFind(int n) {
    id = new int[n];
    rank = new int[n];
    for (int i = 0; i < n; ++i)
      id[i] = i;
  }

  public void unionByRank(int u, int v) {
    final int i = find(u);
    final int j = find(v);
    if (i == j)
      return;
    if (rank[i] < rank[j]) {
      id[i] = j;
    } else if (rank[i] > rank[j]) {
      id[j] = i;
    } else {
      id[i] = j;
      ++rank[j];
    }
  }

  public boolean connected(int u, int v) {
    return find(u) == find(v);
  }

  public void reset(int u) {
    id[u] = u;
  }

  private int[] id;
  private int[] rank;

  private int find(int u) {
    return id[u] == u ? u : (id[u] = find(id[u]));
  }
}

class Solution {
  public List<Integer> findAllPeople(int n, int[][] meetings, int firstPerson) {
    List<Integer> ans = new ArrayList<>();
    UnionFind uf = new UnionFind(n);
    TreeMap<Integer, List<Pair<Integer, Integer>>> timeToPairs = new TreeMap<>();

    uf.unionByRank(0, firstPerson);

    for (int[] meeting : meetings) {
      final int x = meeting[0];
      final int y = meeting[1];
      final int time = meeting[2];
      timeToPairs.putIfAbsent(time, new ArrayList<>());
      timeToPairs.get(time).add(new Pair<>(x, y));
    }

    for (List<Pair<Integer, Integer>> pairs : timeToPairs.values()) {
      Set<Integer> peopleUnioned = new HashSet<>();
      for (Pair<Integer, Integer> pair : pairs) {
        final int x = pair.getKey();
        final int y = pair.getValue();
        uf.unionByRank(x, y);
        peopleUnioned.add(x);
        peopleUnioned.add(y);
      }
      for (final int person : peopleUnioned)
        if (!uf.connected(person, 0))
          uf.reset(person);
    }

    for (int i = 0; i < n; ++i)
      if (uf.connected(i, 0))
        ans.add(i);

    return ans;
  }
}


C++ solution:
class UnionFind {
 public:
  UnionFind(int n) : id(n), rank(n) {
    iota(id.begin(), id.end(), 0);
  }

  void unionByRank(int u, int v) {
    const int i = find(u);
    const int j = find(v);
    if (i == j)
      return;
    if (rank[i] < rank[j]) {
      id[i] = j;
    } else if (rank[i] > rank[j]) {
      id[j] = i;
    } else {
      id[i] = j;
      ++rank[j];
    }
  }

  bool connected(int u, int v) {
    return find(u) == find(v);
  }

  void reset(int u) {
    id[u] = u;
  }

 private:
  vector<int> id;
  vector<int> rank;

  int find(int u) {
    return id[u] == u ? u : id[u] = find(id[u]);
  }
};

class Solution {
 public:
  vector<int> findAllPeople(int n, vector<vector<int>>& meetings,
                            int firstPerson) {
    vector<int> ans;
    UnionFind uf(n);
    map<int, vector<pair<int, int>>> timeToPairs;

    uf.unionByRank(0, firstPerson);

    for (const vector<int>& meeting : meetings) {
      const int x = meeting[0];
      const int y = meeting[1];
      const int time = meeting[2];
      timeToPairs[time].push_back({x, y});
    }

    for (const auto& [_, pairs] : timeToPairs) {
      unordered_set<int> peopleUnioned;
      for (const auto& [x, y] : pairs) {
        uf.unionByRank(x, y);
        peopleUnioned.insert(x);
        peopleUnioned.insert(y);
      }
      for (const int person : peopleUnioned)
        if (!uf.connected(person, 0))
          uf.reset(person);
    }

    for (int i = 0; i < n; ++i)
      if (uf.connected(i, 0))
        ans.push_back(i);

    return ans;
  }
};