> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Find All People With Secret Python Solution

> Tested Python solution for LeetCode 2092 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 2092, [Hard](/catalog/hard). Topics: [Depth-First Search](/catalog/topics/depth-first-search), [Breadth-First Search](/catalog/topics/breadth-first-search), [Union-Find](/catalog/topics/union-find), [Graph Theory](/catalog/topics/graph-theory), [Sorting](/catalog/topics/sorting). [View on LeetCode](https://leetcode.com/problems/find-all-people-with-secret/description/).

Generate this problem as a practice environment: tested reference solution, 16 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 2092   # by problem number
lcpy gen -s find_all_people_with_secret   # by problem name
```

## Problem

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**.

### Examples

```
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.
```

### Constraints

* `2 <= n <= 10^5`
* `1 <= meetings.length <= 10^5`
* `meetings[i].length == 3`
* `0 <= xi, yi <= n - 1`
* `xi != yi`
* `1 <= timei <= 10^5`
* `1 <= firstPerson <= n - 1`

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/find_all_people_with_secret/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/find_all_people_with_secret/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(m log m) for sorting meetings plus near-linear union-find passes
    # Space: O(n) for the parent array
    def find_all_people(self, n: int, meetings: list[list[int]], first_person: int) -> list[int]:
        parent = list(range(n))

        def find(node: int) -> int:
            while parent[node] != node:
                parent[node] = parent[parent[node]]
                node = parent[node]
            return node

        def union(a: int, b: int) -> None:
            root_a, root_b = find(a), find(b)
            if root_a != root_b:
                parent[root_a] = root_b

        known = {0, first_person}
        sorted_meetings = sorted(meetings, key=lambda meeting: meeting[2])
        i = 0
        total = len(sorted_meetings)
        while i < total:
            time = sorted_meetings[i][2]
            participants: set[int] = set()
            while i < total and sorted_meetings[i][2] == time:
                x, y, _ = sorted_meetings[i]
                union(x, y)
                participants.update((x, y))
                i += 1
            knower_roots = {find(p) for p in participants if p in known}
            for person in participants:
                if find(person) in knower_roots:
                    known.add(person)
            for person in participants:
                parent[person] = person
        return list(known)
```

## Complexity

| Time | Space |
| - | - |
| O(m log m) for sorting meetings plus near-linear union-find passes | O(n) for the parent array |

## Tags

[NeetCode All](/catalog/neetcode).


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