Leetcode #2092: Find All People With Secret
In this guide, we solve Leetcode #2092 Find All People With Secret in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Graph, Sorting
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
class Solution:
def findAllPeople(
self, n: int, meetings: List[List[int]], firstPerson: int
) -> List[int]:
vis = [False] * n
vis[0] = vis[firstPerson] = True
meetings.sort(key=lambda x: x[2])
i, m = 0, len(meetings)
while i < m:
j = i
while j + 1 < m and meetings[j + 1][2] == meetings[i][2]:
j += 1
s = set()
g = defaultdict(list)
for x, y, _ in meetings[i : j + 1]:
g[x].append(y)
g[y].append(x)
s.update([x, y])
q = deque([u for u in s if vis[u]])
while q:
u = q.popleft()
for v in g[u]:
if not vis[v]:
vis[v] = True
q.append(v)
i = j + 1
return [i for i, v in enumerate(vis) if v]
Complexity
The time complexity is , and the space complexity is , where and are the number of meetings and the number of experts, respectively. The space complexity is , where and are the number of meetings and the number of experts, respectively.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.