最小未被占据椅子的编号

/* Problem statement: 
There are n friends attending a party, each arriving and leaving at specific times. Each friend will occupy the smallest numbered unoccupied chair upon arrival. We need to determine the chair number occupied by the target friend.

Solution using Java Stream API:
1. Sort the friends by their arrival times using Stream.
2. Use a TreeSet to keep track of the available chairs.
3. Iterate over sorted events and manage chair allocation and deallocation.
*/

import java.util.*;
import java.util.stream.*;

public class PartyChairsStream {
    public int smallestChair(int[][] times, int targetFriend) {
        int n = times.length;

        // List of events (arrival or leaving)
        List<int[]> events = IntStream.range(0, n)
            .mapToObj(i -> new int[][]{{times[i][0], i, 1}, {times[i][1], i, -1}})
            .flatMap(Arrays::stream)
            .sorted((a, b) -> a[0] == b[0] ? a[2] - b[2] : a[0] - b[0])
            .collect(Collectors.toList());

        // TreeSet to manage available chairs
        TreeSet<Integer> availableChairs = IntStream.range(0, n).boxed().collect(Collectors.toCollection(TreeSet::new));
        Map<Integer, Integer> occupiedChairs = new HashMap<>();

        for (int[] event : events) {
            int time = event[0];
            int friend = event[1];
            int type = event[2];

            if (type == 1) { // arrival
                int chair = availableChairs.pollFirst();
                occupiedChairs.put(friend, chair);
                if (friend == targetFriend) {
                    return chair;
                }
            } else { // leaving
                int chair = occupiedChairs.remove(friend);
                availableChairs.add(chair);
            }
        }

        return -1; // should never reach here
    }
}