cpp-flashcards/org/study_deck_02/dsa/graphs/2092-find-all-people-with-secret.org

#+ANKI_DECK: study_deck_02
* TODO 2092. Find All People With Secret :hard:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*2092. Find All People With Secret][2092. Find All People With Secret]]
:END:

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] = [x_{i}, y_{i}, time_{i}]~ indicates that person ~x_{i}~ and person ~y_{i}~ have a meeting at ~time_{i}~. 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 ~x_{i}~ has the secret at ~time_{i}~, then they will share the secret with person ~y_{i}~, 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:*


#+begin_src
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.
#+end_src


*Example 2:*


#+begin_src
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.
#+end_src


*Example 3:*


#+begin_src
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.
#+end_src


*Constraints:*

 - ~2 <= n <= 10^{5}~

 - ~1 <= meetings.length <= 10^{5}~

 - ~meetings[i].length == 3~

 - ~0 <= x_{i}, y_{i }<= n - 1~

 - ~x_{i} != y_{i}~

 - ~1 <= time_{i} <= 10^{5}~

 - ~1 <= firstPerson <= n - 1~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class Solution:
    def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
#+end_src

** TODO C++
#+begin_src cpp
class Solution {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        
    }
};
#+end_src