0055_P2092_Find_All_People_With_Secret.cpp

/* 
Day: 55
Problem Number: 2092 (https://leetcode.com/problems/find-all-people-with-secret)
Date: 24-02-2024 
Description:
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 <= 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

Code:  */
class Solution {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        vector<bool> vis(n);
        vis[0] = vis[firstPerson] = true;
        sort(meetings.begin(), meetings.end(), [&](const auto& x, const auto& y) {
            return x[2] < y[2];
        });
        for (int i = 0, m = meetings.size(); i < m;) {
            int j = i;
            for (; j + 1 < m && meetings[j + 1][2] == meetings[i][2]; ++j)
                ;
            unordered_map<int, vector<int>> g;
            unordered_set<int> s;
            for (int k = i; k <= j; ++k) {
                int x = meetings[k][0], y = meetings[k][1];
                g[x].push_back(y);
                g[y].push_back(x);
                s.insert(x);
                s.insert(y);
            }
            queue<int> q;
            for (int u : s)
                if (vis[u])
                    q.push(u);
            while (!q.empty()) {
                int u = q.front();
                q.pop();
                for (int v : g[u]) {
                    if (!vis[v]) {
                        vis[v] = true;
                        q.push(v);
                    }
                }
            }
            i = j + 1;
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i)
            if (vis[i])
                ans.push_back(i);
        return ans;
    }
};