Find All People With Secret
Tags: leetcode, dsa, theabbie
Published: 2022-06-03 10:30:41

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:**

    from collections import defaultdict
    
    class Solution:
        def getParent(self, parent, x):
            if parent[x] == x:
                return x
            parent[x] = self.getParent(parent, parent[x])
            return parent[x]
        
        def connect(self, parent, a, b):
            parent[self.getParent(parent, a)] = self.getParent(parent, b)
            
        def isConnected(self, parent, a, b):
            return self.getParent(parent, a) == self.getParent(parent, b)
        
        def findAllPeople(self, n: int, meetings: List[List[int]], firstPerson: int) -> List[int]:
            parent = list(range(n))
            meets = defaultdict(list)
            for a, b, t in meetings:
                meets[t].append((a, b))
            self.connect(parent, 0, firstPerson)
            people = set()
            for t in sorted(meets.keys()):
                people.clear()
                for a, b in meets[t]:
                    self.connect(parent, a, b)
                    people.update({a, b})
                for p in people:
                    if not self.isConnected(parent, p, 0):
                        parent[p] = p
            return [i for i in range(n) if self.isConnected(parent, i, 0)]