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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2001_2100.s2092_find_all_people_with_secret;

// #Hard #Sorting #Depth_First_Search #Breadth_First_Search #Graph #Union_Find
// #2022_05_27_Time_95_ms_(84.86%)_Space_99.7_MB_(87.33%)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

/**
 * 2092 - Find All People With Secret\.
 *
 * Hard
 *
 * 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:**
 *
 * 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 <= x_i, y_i <= n - 1
 * *   x_i != y_i
 * *   1 <= time_i <= 10⁵
 * *   `1 <= firstPerson <= n - 1`
**/
public class Solution {
    public List findAllPeople(int n, int[][] meetings, int firstPerson) {
        Arrays.sort(meetings, ((a, b) -> a[2] - b[2]));
        UF uf = new UF(n);
        // base
        uf.union(0, firstPerson);
        // for every time we have a pool of people that talk to each other
        // if someone knows a secret proir to this meeting - all pool will too
        // if not - reset unions from this pool
        int i = 0;
        while (i < meetings.length) {
            int curTime = meetings[i][2];
            Set pool = new HashSet<>();
            while (i < meetings.length && curTime == meetings[i][2]) {
                int[] currentMeeting = meetings[i];
                uf.union(currentMeeting[0], currentMeeting[1]);
                pool.add(currentMeeting[0]);
                pool.add(currentMeeting[1]);
                i++;
            }
            // meeting that took place now should't affect future
            // meetings if people don't know the secret
            for (int j : pool) {
                if (!uf.connected(0, j)) {
                    uf.reset(j);
                }
            }
        }
        // if the person is conneted to 0 - they know a secret
        List ans = new ArrayList<>();
        for (int j = 0; j < n; j++) {
            if (uf.connected(j, 0)) {
                ans.add(j);
            }
        }
        return ans;
    }

    // regular union find
    private static class UF {
        private int[] parent;
        private int[] rank;

        public UF(int n) {
            parent = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }

        public void union(int p, int q) {
            int rootP = find(p);
            int rootQ = find(q);
            if (rootP == rootQ) {
                return;
            }
            if (rank[rootP] < rank[rootQ]) {
                parent[rootP] = rootQ;
            } else {
                parent[rootQ] = rootP;
                rank[rootP]++;
            }
        }

        public int find(int p) {
            while (parent[p] != p) {
                p = parent[parent[p]];
            }
            return p;
        }

        public boolean connected(int p, int q) {
            return find(p) == find(q);
        }

        public void reset(int p) {
            parent[p] = p;
            rank[p] = 0;
        }
    }
}