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Find All People With Secret⚓︎

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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] = [x_i, y_i, time_i] indicates that person x_i and person y_i have a meeting at time_i. 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 x_i has the secret at time_i, then they will share the secret with person y_i, 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:
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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:

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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:

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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 <= x_i, y_i <= n - 1
  • x_i != y_i
  • 1 <= time_i <= 10^5
  • 1 <= firstPerson <= n - 1

Solution⚓︎

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class Solution {
private:
    vector<int> p;
    vector<bool> secret;

    void build(int n, int firstPerson) {
        p.resize(n);
        secret.resize(n);
        for (int i = 0; i < n; i++) {
            p[i] = i;
            secret[i] = false;
        }

        p[firstPerson] = 0;
        secret[0] = true;
    }

    int find(int i) {
        if (i != p[i]) p[i] = find(p[i]);
        return p[i];
    }

    void unionSets(int x, int y) {
        int fx = find(x), fy = find(y);
        if (fx != fy) {
            p[fx] = fy;
            secret[fy] = secret[fx] || secret[fy];
        }
    }

public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        build(n, firstPerson);
        sort(meetings.begin(), meetings.end(), [](auto& lhs, auto& rhs) { 
            return lhs[2] < rhs[2]; 
        });

        int m = meetings.size();
        for (int l = 0, r; l < m; ) {
            r = l;
            while (r + 1 < m && meetings[l][2] == meetings[r + 1][2]) {
                r++;
            }

            for (int i = l; i <= r; i++) {
                unionSets(meetings[i][0], meetings[i][1]);
            }

            for (int i = l, a, b; i <= r; i++) {
                a = meetings[i][0]; b = meetings[i][1];
                // reset
                if (!secret[find(a)]) p[a] = a;
                if (!secret[find(b)]) p[b] = b;
            }

            l = r + 1;
        }

        vector<int> res;
        for (int i = 0; i < n; i++) {
            if (secret[find(i)]) res.emplace_back(i);
        }
        return res;
    }
};
  • Time complexity: \(O(m\log m + m + n)\);
  • Space complexity: \(O(n)\).