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2213. Find All People With Secret

Depth-First Search Breadth-First Search Union Find Graph Sorting

Problem - Find All People With Secret

Hard

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

Solutions

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class Solution:
    def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
        visited = [False] * n
        visited[0] = visited[firstPerson] = True
        meetings.sort(key=lambda x: x[2])
        i, m = 0, len(meetings)
        while m > i:
            j = i
            while j + 1 < m and meetings[j + 1][2] == meetings[i][2]:
                j += 1
            s = set()
            graph = defaultdict(list)
            for x, y, _ in meetings[i : j + 1]:
                graph[x].append(y)
                graph[y].append(x)
                s.update([x, y])
            queue = deque([u for u in s if visited[u]])
            while queue:
                u = queue.popleft()
                for v in graph[u]:
                    if not visited[v]:
                        visited[v] = True
                        queue.append(v)
            i = j + 1

        return [i for i, val in enumerate(visited) if val]

Submission Stats:

  • Runtime: 386 ms (36.31%)
  • Memory: 61.1 MB (68.79%)