Exam Room

Problem

There is an exam room with n seats in a single row labeled from 0 to n -1.

When a student enters the room, they must sit in the seat that maximizes the distance to the closest person. If there are multiple such seats, they sit in the seat with the lowest number. If no one is in the room, then the student sits at seat number 0.

Design a class that simulates the mentioned exam room.

Implement the ExamRoom class:

• ExamRoom(int n) Initializes the object of the exam room with the number of the seats n.
• int seat() Returns the label of the seat at which the next student will set.
• void leave(int p) Indicates that the student sitting at seat p will leave the room. It is guaranteed that there will be a student sitting at seat p.

Examples

Example 1

**Input**
["ExamRoom", "seat", "seat", "seat", "seat", "leave", "seat"]
[[10], [], [], [], [], [4], []]
**Output**
[null, 0, 9, 4, 2, null, 5]

**Explanation**
ExamRoom examRoom = new ExamRoom(10);
examRoom.seat(); // return 0, no one is in the room, then the student sits at seat number 0.
examRoom.seat(); // return 9, the student sits at the last seat number 9.
examRoom.seat(); // return 4, the student sits at the last seat number 4.
examRoom.seat(); // return 2, the student sits at the last seat number 2.
examRoom.leave(4);
examRoom.seat(); // return 5, the student sits at the last seat number 5.

Constraints

• 1 <= n <= 10^9
• It is guaranteed that there is a student sitting at seat p.
• At most 104 calls will be made to seat and leave.

Solution

Method 1 – Ordered Set and Interval Greedy

Intuition

To maximize the distance to the closest person, always choose the largest empty interval. Use an ordered set to keep track of occupied seats and compute the best seat each time. For leaving, simply remove the seat from the set.

Approach

1. Use a sorted set to store occupied seats.
2. For seat(), check the distance from 0 to the first seat, between all pairs of occupied seats, and from the last seat to n-1. Choose the seat that maximizes the minimum distance.
3. For leave(p), remove p from the set.

Code

C++

class ExamRoom {
    set<int> seats;
    int n;
public:
    ExamRoom(int n): n(n) {}
    int seat() {
        if (seats.empty()) {
            seats.insert(0);
            return 0;
        }
        int prev = -1, max_dist = 0, ans = 0;
        for (int s : seats) {
            if (prev == -1) {
                if (s > max_dist) {
                    max_dist = s;
                    ans = 0;
                }
            } else {
                int d = (s - prev) / 2;
                if (d > max_dist) {
                    max_dist = d;
                    ans = prev + d;
                }
            }
            prev = s;
        }
        if (n - 1 - *seats.rbegin() > max_dist) {
            ans = n - 1;
        }
        seats.insert(ans);
        return ans;
    }
    void leave(int p) {
        seats.erase(p);
    }
};

Java

import java.util.TreeSet;
class ExamRoom {
    TreeSet<Integer> seats = new TreeSet<>();
    int n;
    public ExamRoom(int n) { this.n = n; }
    public int seat() {
        if (seats.isEmpty()) {
            seats.add(0);
            return 0;
        }
        int prev = -1, maxDist = 0, ans = 0;
        for (int s : seats) {
            if (prev == -1) {
                if (s > maxDist) {
                    maxDist = s;
                    ans = 0;
                }
            } else {
                int d = (s - prev) / 2;
                if (d > maxDist) {
                    maxDist = d;
                    ans = prev + d;
                }
            }
            prev = s;
        }
        if (n - 1 - seats.last() > maxDist) {
            ans = n - 1;
        }
        seats.add(ans);
        return ans;
    }
    public void leave(int p) {
        seats.remove(p);
    }
}

Python

class ExamRoom:
    def __init__(self, n: int):
        self.n = n
        self.seats = set()
    def seat(self) -> int:
        if not self.seats:
            self.seats.add(0)
            return 0
        arr = sorted(self.seats)
        max_dist = arr[0]
        ans = 0
        for i in range(1, len(arr)):
            d = (arr[i] - arr[i-1]) // 2
            if d > max_dist:
                max_dist = d
                ans = arr[i-1] + d
        if self.n - 1 - arr[-1] > max_dist:
            ans = self.n - 1
        self.seats.add(ans)
        return ans
    def leave(self, p: int) -> None:
        self.seats.remove(p)

Rust

use std::collections::BTreeSet;
struct ExamRoom {
    n: i32,
    seats: BTreeSet<i32>,
}
impl ExamRoom {
    fn new(n: i32) -> Self {
        Self { n, seats: BTreeSet::new() }
    }
    fn seat(&mut self) -> i32 {
        if self.seats.is_empty() {
            self.seats.insert(0);
            return 0;
        }
        let mut prev = -1;
        let mut max_dist = 0;
        let mut ans = 0;
        for &s in &self.seats {
            if prev == -1 {
                if s > max_dist {
                    max_dist = s;
                    ans = 0;
                }
            } else {
                let d = (s - prev) / 2;
                if d > max_dist {
                    max_dist = d;
                    ans = prev + d;
                }
            }
            prev = s;
        }
        if self.n - 1 - *self.seats.iter().rev().next().unwrap() > max_dist {
            ans = self.n - 1;
        }
        self.seats.insert(ans);
        ans
    }
    fn leave(&mut self, p: i32) {
        self.seats.remove(&p);
    }
}

TypeScript

class ExamRoom {
    private n: number;
    private seats: Set<number>;
    constructor(n: number) {
        this.n = n;
        this.seats = new Set();
    }
    seat(): number {
        if (this.seats.size === 0) {
            this.seats.add(0);
            return 0;
        }
        const arr = Array.from(this.seats).sort((a, b) => a - b);
        let maxDist = arr[0], ans = 0;
        for (let i = 1; i < arr.length; ++i) {
            const d = Math.floor((arr[i] - arr[i-1]) / 2);
            if (d > maxDist) {
                maxDist = d;
                ans = arr[i-1] + d;
            }
        }
        if (this.n - 1 - arr[arr.length-1] > maxDist) {
            ans = this.n - 1;
        }
        this.seats.add(ans);
        return ans;
    }
    leave(p: number): void {
        this.seats.delete(p);
    }
}

Complexity

• ⏰ Time complexity: O(log n) per seat/leave operation (for ordered set operations).
• 🧺 Space complexity: O(n) for storing occupied seats.