2092. Find All People With Secret

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 <= 105
• 1 <= meetings.length <= 105
• meetings[i].length == 3
• 0 <= xi, yi <= n - 1
• xi != yi
• 1 <= timei <= 105
• 1 <= firstPerson <= n - 1

 

Solutions

Solution: Union Find

• Time complexity: O(mlogm+n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} meetings
 * @param {number} firstPerson
 * @return {number[]}
 */
const findAllPeople = function (n, meetings, firstPerson) {
  const uf = new UnionFind(n);
  const meetingMap = new Map();
  const result = [];

  uf.union(0, firstPerson);
  meetings.sort((a, b) => a[2] - b[2]);

  for (const [x, y, time] of meetings) {
    if (!meetingMap.has(time)) {
      meetingMap.set(time, []);
    }

    meetingMap.get(time).push({ x, y });
  }

  for (const paris of meetingMap.values()) {
    const peoples = new Set();

    for (const { x, y } of paris) {
      uf.union(x, y);
      peoples.add(x);
      peoples.add(y);
    }

    for (const people of peoples) {
      if (uf.find(people) !== uf.find(0)) {
        uf.reset(people);
      }
    }
  }

  for (let index = 0; index < n; index++) {
    if (uf.find(index) === uf.find(0)) {
      result.push(index);
    }
  }

  return result;
};

class UnionFind {
  constructor(n) {
    this.groups = Array.from({ length: n }, (_, index) => index);
    this.ranks = Array.from({ length: n }, () => 0);
  }

  find(x) {
    if (this.groups[x] === x) return x;

    this.groups[x] = this.find(this.groups[x]);

    return this.groups[x];
  }

  union(x, y) {
    const groupX = this.find(x);
    const groupY = this.find(y);

    if (groupX === groupY) return false;
    if (this.ranks[groupX] > this.ranks[groupY]) {
      this.groups[groupY] = groupX;
    } else if (this.ranks[groupX] < this.ranks[groupY]) {
      this.groups[groupX] = groupY;
    } else {
      this.groups[groupY] = groupX;
      this.ranks[groupX] += 1;
    }

    return true;
  }

  reset(x) {
    this.groups[x] = x;
  }
}