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Abhishek Chaudhary

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# 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:

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)]
``````