1942. The Number of the Smallest Unoccupied Chair

Problem

There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity. When a friend arrives at the party, they sit on the unoccupied chair with the smallest number.

• For example, if chairs 0, 1, and 5 are occupied when a friend comes, they will sit on chair number 2.

When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment, they can sit in that chair.

You are given a 0-indexed 2D integer array times where times[i] = [arrivali, leavingi], indicating the arrival and leaving times of the ith friend respectively, and an integer targetFriend. All arrival times are distinct.

Return** the chair number that the friend numbered targetFriend will sit on**.

  Example 1:

Input: times = [[1,4],[2,3],[4,6]], targetFriend = 1
Output: 1
Explanation: 
- Friend 0 arrives at time 1 and sits on chair 0.
- Friend 1 arrives at time 2 and sits on chair 1.
- Friend 1 leaves at time 3 and chair 1 becomes empty.
- Friend 0 leaves at time 4 and chair 0 becomes empty.
- Friend 2 arrives at time 4 and sits on chair 0.
Since friend 1 sat on chair 1, we return 1.

Example 2:

Input: times = [[3,10],[1,5],[2,6]], targetFriend = 0
Output: 2
Explanation: 
- Friend 1 arrives at time 1 and sits on chair 0.
- Friend 2 arrives at time 2 and sits on chair 1.
- Friend 0 arrives at time 3 and sits on chair 2.
- Friend 1 leaves at time 5 and chair 0 becomes empty.
- Friend 2 leaves at time 6 and chair 1 becomes empty.
- Friend 0 leaves at time 10 and chair 2 becomes empty.
Since friend 0 sat on chair 2, we return 2.

  Constraints:

• n == times.length

• 2 <= n <= 10^4

• times[i].length == 2

• 1 <= arrivali < leavingi <= 10^5

• 0 <= targetFriend <= n - 1

• Each arrivali time is distinct.

Solution (Java)

class Solution {
    public int smallestChair(int[][] times, int targetFriend) {
        PriorityQueue<Integer> minheap = new PriorityQueue<>();
        minheap.offer(0);
        Person[] all = new Person[times.length * 2];
        for (int i = 0; i < times.length; i++) {
            all[2 * i] = new Person(i, times[i][0], false, true);
            all[2 * i + 1] = new Person(i, times[i][1], true, false);
        }
        Arrays.sort(
                all,
                (a, b) -> {
                    int i = a.leave ? -1 : 1;
                    int j = b.leave ? -1 : 1;
                    return a.time == b.time ? i - j : a.time - b.time;
                });

        int[] seat = new int[times.length];
        for (int i = 0; true; i++) {
            if (all[i].arrive) {
                if (targetFriend == all[i].idx) {
                    return minheap.peek();
                }
                seat[all[i].idx] = minheap.poll();
                if (minheap.isEmpty()) {
                    minheap.offer(seat[all[i].idx] + 1);
                }
            } else {
                minheap.offer(seat[all[i].idx]);
            }
        }
    }

    private static class Person {
        boolean leave;
        boolean arrive;
        int time;
        int idx;

        Person(int idx, int time, boolean leave, boolean arrive) {
            this.time = time;
            this.leave = leave;
            this.arrive = arrive;
            this.idx = idx;
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).