彼此熟识的最早时间

/*
 * LeetCode 1101: The Earliest Moment When Everyone Become Friends
 * 
 * Problem statement: Given a list of logs, where each log entry is a triplet (timestamp, person1, person2),
 * representing that person1 and person2 became friends at that timestamp. We are to find the earliest
 * time at which all people became friends. If there is no such time, return -1.
 * 
 * Approach using Java Streams: 
 * 1. We will use the union-find (disjoint-set) data structure to keep track of connected components (groups of friends).
 * 2. Use streams to sort the logs based on the timestamps.
 * 3. Iterate over the sorted logs and union the sets of the two persons in each log.
 * 4. After each union operation, check if all persons are in the same connected component.
 * 5. If they are, return the current timestamp. If we finish processing all logs and they are not all connected, return -1.
 */

import java.util.Arrays;
import java.util.Comparator;
import java.util.stream.Stream;

public class Solution {
    public int earliestAcq(int[][] logs, int N) {
        UnionFind uf = new UnionFind(N);

        return Arrays.stream(logs) // Step 2
            .sorted(Comparator.comparingInt(log -> log[0]))
            .filter(log -> {
                uf.union(log[1], log[2]);
                return uf.getCount() == 1; // Step 4
            })
            .findFirst() // Step 3
            .map(log -> log[0]) // Step 5
            .orElse(-1); // Step 5
    }

    class UnionFind {
        private int[] parent;
        private int[] rank;
        private int count;

        public UnionFind(int n) {
            parent = new int[n];
            rank = new int[n];
            count = n;
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                rank[i] = 1;
            }
        }

        public int find(int p) {
            if (p != parent[p]) {
                parent[p] = find(parent[p]); // Path compression
            }
            return parent[p];
        }

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

        public int getCount() {
            return count;
        }
    }
}