将区间分为最少组数

/*
 * Problem Statement:
 * Given a list of intervals, group them such that no two intervals in the same group overlap.
 * Find the minimum number of groups required.
 * 
 * Approach:
 * 1. Sort intervals using Java Streams.
 * 2. Utilize a PriorityQueue to manage group end times.
 * 3. For each interval, if it doesn't overlap with the earliest ending interval, reuse that group.
 * 4. Otherwise, create a new group.
 */
import java.util.*;
import java.util.stream.Collectors;

public class IntervalPartitionStream {
    public int minGroups(int[][] intervals) {
        // Sort intervals using streams
        List<int[]> sortedIntervals = Arrays.stream(intervals)
            .sorted((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0])
            .collect(Collectors.toList());
        // PriorityQueue to manage group end times
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int[] interval : sortedIntervals) {
            if (!pq.isEmpty() && pq.peek() < interval[0]) {
                pq.poll();
            }
            pq.add(interval[1]);
        }
        return pq.size();
    }
}