Find All People With Secret Problem Statement

There are n people numbered from 0 to n - 1. A list of meetings is given whose each entry is a tuple [xi, yi, timei], indicating that the person xi and the person yi have a meeting at timei.
Person 0 has a secret and initially shares the secret with a given person at time 0. This secret is then shared every time a meeting takes place with the person that has the secret.
Return a list of all the people who have the secret after all the meetings have taken place.

Example One

{
"n": 6,
"meetings": [
[1, 2, 3],
[2, 3, 5],
[1, 5, 2],
[3, 5, 10]
],
"first_person": 2
}

Output:

[0, 1, 2, 3, 5]

At time 0, person 0 shares the secret with person 2.
At time 2, neither person 1 nor person 5 knows the secret.
At time 3, person 2 shares the secret with person 1.
At time 5, person 2 shares the secret with person 3.
At time 10, person 3 shares the secret with person 5.
Thus, people 0, 1, 2, 3, and 5 know the secret after all the meetings.

Example Two

{
"n": 5,
"meetings": [
[2, 3, 1],
[2, 4, 1],
[1, 2, 1]
],
"first_person": 1
}

Output:

[0, 1, 2, 3, 4]

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 persons 3 and 4.
Thus, people 0, 1, 2, 3, and 4 know the secret after all the meetings.

Notes

• A person may attend multiple meetings at the same time.
• The secrets are shared instantaneously, which means a person may receive the secret and share it with people in other meetings within the same time frame.
• Return the output list in any order.

Constraints:

• 2 <= n <= 10⁵
• 0 <= size of the input list <= 10⁵
• 0 <= xi, yi <= n – 1
• xi != yi
• 1 <= timei <= 10⁵
• 1 <= first person with whom secret is shared at time 0 <= n – 1

Code For Find All People With Secret Solution: Bfs

/*
Asymptotic complexity in terms of the number of meetings `m` and the number of people `n`:
* Time: O(m * log(m) + n).
* Auxiliary space: O(n + m).
* Total space: O(n + m).
*/
vector<int> find_all_people_with_secret(int n, vector<vector<int>> &meetings, int first_person) {
    vector<vector<pair<int,int>>> graph(n);
    for (int i = 0; i < meetings.size(); i++) {
        graph[meetings[i][0]].push_back({meetings[i][1], meetings[i][2]});
        graph[meetings[i][1]].push_back({meetings[i][0], meetings[i][2]});
    }

    // Min Heap which stores a pair of elements where the first entry denotes the current
    // time and the second entry denotes the person who knows the secret.
    priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
    vector<int> is_secret_known(n, false);

    pq.push({0, 0});
    pq.push({0, first_person});

    while (!pq.empty()) {
        int current_time = pq.top().first;
        int person_x = pq.top().second;
        pq.pop();

        if (is_secret_known[person_x]) {
            continue;
        }

        is_secret_known[person_x] = true;

        for (int i = 0; i < graph[person_x].size(); i++) {
            int person_y = graph[person_x][i].first;
            int meeting_time = graph[person_x][i].second;
            if (is_secret_known[person_y]) {
                continue;
            }
            if (meeting_time >= current_time) {
                pq.push({meeting_time, person_y});
            }
        }
    }

    vector<int> people;
    for (int i = 0; i < n; i++) {
        if (is_secret_known[i]) {
            people.push_back(i);
        }
    }

    return people;
}

We hope that these solutions to the 4 sum problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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