2213 Find All People With Secret
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.
Example 1:
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**Input:** n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1
**Output:** [0,1,2,3,5]
Explanation:
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.
Example 2:
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**Input:** n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3
**Output:** [0,1,3]
**Explanation:**
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.
Example 3:
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**Input:** n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1
**Output:** [0,1,2,3,4]
**Explanation:**
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.
Constraints:
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2 <= n <= 105
1 <= meetings.length <= 105
meetings[i].length == 3
0 <= xi, yi <= n - 1
xi != yi
1 <= timei <= 105
1 <= firstPerson <= n - 1
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class Solution:
def findAllPeople(
self, n: int, meetings: List[List[int]], firstPerson: int
) -> List[int]:
# Sort meetings in increasing order of time
meetings.sort(key=lambda x: x[2])
# Group Meetings in increasing order of time
same_time_meetings = defaultdict(list)
for x, y, t in meetings:
same_time_meetings[t].append((x, y))
# Create graph
graph = UnionFind(n)
graph.unite(firstPerson, 0)
# Process in increasing order of time
for t in same_time_meetings:
# Unite two persons taking part in a meeting
for x, y in same_time_meetings[t]:
graph.unite(x, y)
# If any one knows the secret, both will be connected to 0.
# If no one knows the secret, then reset.
for x, y in same_time_meetings[t]:
if not graph.connected(x, 0):
# No need to check for y since x and y were united
graph.reset(x)
graph.reset(y)
# Al those who are connected to 0 will know the secret
return [i for i in range(n) if graph.connected(i, 0)]
class UnionFind:
def __init__(self, nodes: int):
# Initialize parent and rank arrays
self.parent = [i for i in range(nodes)]
self.rank = [0] * nodes
def find(self, x: int) -> int:
# Find the parent of node x. Use Path Compression
if self.parent[x] != x:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def unite(self, x: int, y: int) -> None:
# Unite two nodes x and y, if they are not already united
px = self.find(x)
py = self.find(y)
if px != py:
# Union by Rank Heuristic
if self.rank[px] > self.rank[py]:
self.parent[py] = px
elif self.rank[px] < self.rank[py]:
self.parent[px] = py
else:
self.parent[py] = px
self.rank[px] += 1
def connected(self, x: int, y: int) -> bool:
# Check if two nodes x and y are connected or not
return self.find(x) == self.find(y)
def reset(self, x: int) -> None:
# Reset the initial properties of node x
self.parent[x] = x
self.rank[x] = 0