安排会议日程

/*
 * Leetcode 1229: Meeting Scheduler using Java Streams
 * Given the availability time slots arrays slots1 and slots2 of two people and a meeting duration, 
 * return the earliest time slot that works for both of them and is of duration duration.
 * If there is no common time slot that satisfies the duration requirement, return an empty array.
 *
 * Approach:
 * 1. Convert slots1 and slots2 into sorted lists of int arrays.
 * 2. Use nested streams to find the earliest overlapping slot with the required duration.
 * 3. If a suitable time slot is found, return the start time and start time + duration as a list.
 * 4. If no suitable time slot is found, return an empty list.
 */
import java.util.*;
import java.util.stream.*;

public class MeetingSchedulerStream {
    public List<Integer> minAvailableDuration(int[][] slots1, int[][] slots2, int duration) {
        List<int[]> list1 = Arrays.stream(slots1).sorted(Comparator.comparingInt(a -> a[0])).collect(Collectors.toList());
        List<int[]> list2 = Arrays.stream(slots2).sorted(Comparator.comparingInt(a -> a[0])).collect(Collectors.toList());

        return list1.stream()
            .flatMap(slot1 -> list2.stream()
                .map(slot2 -> new int[] { Math.max(slot1[0], slot2[0]), Math.min(slot1[1], slot2[1]) })
                .filter(interval -> interval[1] - interval[0] >= duration)
                .findFirst()
                .stream())
            .map(interval -> Arrays.asList(interval[0], interval[0] + duration))
            .findFirst()
            .orElse(Collections.emptyList());
    }
}