2092. Find All People With Secret

Problem

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

/**
 * @param {number} n
 * @param {number[][]} meetings
 * @param {number} firstPerson
 * @return {number[]}
 */
var findAllPeople = function(n, meetings, firstPerson) {
    var map = Array(n).fill(0).map(() => []);
    var queue = new MinPriorityQueue();
    var hasSecretMap = Array(n);
    var handledMap = Array(meetings.length);
    hasSecretMap[0] = true;
    hasSecretMap[firstPerson] = true;
    for (var i = 0; i < meetings.length; i++) {
        map[meetings[i][0]].push([meetings[i][1], meetings[i][2], i]);
        map[meetings[i][1]].push([meetings[i][0], meetings[i][2], i]);
        queue.enqueue(
            [...meetings[i], i],
            (hasSecretMap[meetings[i][0]] || hasSecretMap[meetings[i][1]])
                ? meetings[i][2] - 0.1
                : meetings[i][2]
        );
    }
    while (queue.size() !== 0) {
        var item = queue.dequeue().element;
        if (handledMap[item[3]]) continue;
        handledMap[item[3]] = true;
        if (!hasSecretMap[item[0]] && !hasSecretMap[item[1]]) continue;
        if (hasSecretMap[item[0]] && hasSecretMap[item[1]]) continue;
        var num = hasSecretMap[item[0]] ? item[1] : item[0];
        for (var i = 0; i < map[num].length; i++) {
            var data = map[num][i];
            if (handledMap[data[2]]) continue;
            if (hasSecretMap[data[0]]) continue;
            queue.enqueue(
                [num, data[0], data[1] - 0.1, data[2]],
                data[1] - 0.1
            );
        }
        hasSecretMap[num] = true;
    }
    return hasSecretMap.reduce((res, item, i) => {
        if (item) res.push(i);
        return res;
    }, []);
};

Explain:

Priority queue with dynamicly changes priority.

Complexity:

• Time complexity : O(n * log(n) + m).
• Space complexity : O(n + m).