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:
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:
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:
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:
-
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
SOLUTION:
from collections import defaultdict
class Solution:
def getParent(self, parent, x):
if parent[x] == x:
return x
parent[x] = self.getParent(parent, parent[x])
return parent[x]
def connect(self, parent, a, b):
parent[self.getParent(parent, a)] = self.getParent(parent, b)
def isConnected(self, parent, a, b):
return self.getParent(parent, a) == self.getParent(parent, b)
def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
parent = list(range(n))
meets = defaultdict(list)
for a, b, t in meetings:
meets[t].append((a, b))
self.connect(parent, 0, firstPerson)
people = set()
for t in sorted(meets.keys()):
people.clear()
for a, b in meets[t]:
self.connect(parent, a, b)
people.update({a, b})
for p in people:
if not self.isConnected(parent, p, 0):
parent[p] = p
return [i for i in range(n) if self.isConnected(parent, i, 0)]
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