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.
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.
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.
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.
2 <= n <= 1051 <= meetings.length <= 105meetings[i].length == 30 <= xi, yi <= n - 1xi != yi1 <= timei <= 1051 <= firstPerson <= n - 1
classSolution: deffindAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) ->List[int]: meetings.sort(key=lambdam: m[2]) knows= {0, firstPerson} currknows=set() edges= {} foriinrange(len(meetings)): x, y, time=meetings[i] ifxinknowsoryinknows: knows.add(x) knows.add(y) currknows.add(x) currknows.add(y) else: ifxnotinedges: edges[x] = [] ifynotinedges: edges[y] = [] edges[x].append(y) edges[y].append(x) ifi+1==len(meetings) ortime!=meetings[i+1][2]: currknows=list(currknows) whilecurrknows!= []: x=currknows.pop() foryinedges.get(x, []): ifynotinknows: knows.add(y) currknows.append(y) currknows=set() edges= {} returnlist(knows)