Formatted question description: https://leetcode.ca/all/2092.html

# 2092. 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.


Example 4:

Input: n = 6, meetings = [[0,2,1],[1,3,1],[4,5,1]], firstPerson = 1
Output: [0,1,2,3]
Explanation:
At time 0, person 0 shares the secret with person 1.
At time 1, person 0 shares the secret with person 2, and person 1 shares the secret with person 3.
Thus, people 0, 1, 2, and 3 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

Similar Questions:

## Solution 1. Union Find

Sort meetings in ascending order of meeting time.

Visit the meetings happening at the same time together.

We can connect the two persons in the same meeting using a UnionFind.

Tricky point here: After traversing this batch of meetings, we reset the persons who don’t know the secret by checking if he/she is connected to person 0. With UnionFind, reseting means setting id[x] = x.

In the end, we add all the persons who are connected to person 0 into the answer array.

// OJ: https://leetcode.com/problems/find-all-people-with-secret/
// Time: O(MlogM + N) where M is the length of meetings
// Space: O(N)
class UnionFind {
vector<int> id;
public:
UnionFind(int n) : id(n) {
iota(begin(id), end(id), 0);
}
void connect(int a, int b) {
id[find(b)] = find(a);
}
int find(int a) {
return id[a] == a ? a : (id[a] = find(id[a]));
}
bool connected(int a, int b) {
return find(a) == find(b);
}
void reset(int a) { id[a] = a; }
};
class Solution {
public:
vector<int> findAllPeople(int n, vector<vector<int>>& A, int firstPerson) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[2] < b[2]; }); // Sort the meetings in ascending order of meeting time
int i = 0, M = A.size();
UnionFind uf(n);
uf.connect(0, firstPerson); // Connect person 0 with the first person
vector<int> ppl;
while (i < M) {
ppl.clear();
int time = A[i][2];
while (i < M && A[i][2] == time) { // For all the meetings happending at the same time
uf.connect(A[i][0], A[i][1]); // Connect the two persons
ppl.push_back(A[i][0]); // Add both persons into the pool
ppl.push_back(A[i][1]);
++i;
}
for (int n : ppl) { // For each person in the pool, check if he/she's connected with person 0.
if (!uf.connected(0, n)) uf.reset(n); // If not, this person doesn't have secret, reset it.
}
}
vector<int> ans;
for (int j = 0; j < n; ++j) {
if (uf.connected(0, j)) ans.push_back(j); // Push all the persons who are connected with person 0 into answer array
}
return ans;
}
};


## Discuss

https://leetcode.com/problems/find-all-people-with-secret/discuss/1599815/C%2B%2B-Union-Find