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2092. Find All People With Secret

Difficulty Topics

Description

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

find-all-people-with-secret.py
class DSU:
    def __init__(self, n):
        self.graph = list(range(n))

    def find(self, x):
        if self.graph[x] != x:
            self.graph[x] = self.find(self.graph[x])

        return self.graph[x]

    def union(self, x, y):
        ux, uy = self.find(x), self.find(y)
        self.graph[ux] = uy

    def connected(self, x, y):
        return self.find(x) == self.find(y)

    def disconnect(self, x):
        self.graph[x] = x

class Solution:
    def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
        m = len(meetings)
        dsu = DSU(n)
        dsu.union(0, firstPerson)
        meetings.sort(key = lambda x : x[2])
        people = set()

        i = 0
        while i < m:
            people.clear()

            time = meetings[i][2]

            while i < m and meetings[i][2] == time:
                x, y, _ = meetings[i]
                dsu.union(x, y)
                people.add(x)
                people.add(y)
                i += 1

            for p in people:
                if not dsu.connected(0, p):
                    dsu.disconnect(p)

        res = []
        for i in range(n):
            if dsu.connected(0, i):
                res.append(i)

        return res