What is the time complexity of Exam Room?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Exam Room?

This problem uses Design, Heap (Priority Queue), Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Exam Room?

Insight 1: Maintaining a sorted list of occupied seats allows efficient calculation of distances to neighboring students. Insight 2: The maximum distance can occur at the beginning of the row (seat 0), in the middle between two students, or at the end of the row (seat n-1).

Which companies ask LeetCode 855. Exam Room in interviews?

This problem is frequently asked at Quora.

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Exam Room - LeetCode 855 Solution

Exam Room - Complete Solution Guide

Exam Room is LeetCode problem 855, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 seat

Detailed Explanation

The problem asks us to simulate an exam room seating arrangement. The exam room has 'n' seats numbered from 0 to n-1. Students enter the room one at a time and must sit in the seat that maximizes their distance to the closest seated student. If there are multiple such seats, the student chooses the seat with the smallest number. If the room is empty, the student sits in seat 0. We need to implement an `ExamRoom` class with the following methods: `ExamRoom(int n)` to initialize the room with 'n' seats, `seat()` to return the seat number for the next student, and `leave(int p)` to indicate that the student in seat 'p' is leaving.

Solution Approach

The solution uses a sorted list called `students` to store the positions of the seated students. When a student wants to sit, the algorithm iterates through the list, calculates the distance between adjacent students, and finds the maximum distance to a seated student. Special care is taken to handle the edges (seat 0 and seat n-1). The `bisect.insort` method in Python or equivalent methods in Java and C++ are used to maintain the sorted order after a student takes a seat. When a student leaves, the algorithm removes their seat number from the `students` list.

Step-by-Step Algorithm

Step 1: Initialize the ExamRoom with 'n' total seats and an empty list called `students` to track occupied seats.
Step 2: In the `seat()` method, if `students` is empty, return seat 0 and add 0 to the `students` list.
Step 3: If `students` is not empty, calculate the distance from seat 0 to the first student (`students[0]`).
Step 4: Iterate through the `students` list, calculating the distance between each pair of adjacent students (`students[i+1] - students[i]`) and divide by 2 to find the midpoint distance.
Step 5: Calculate the distance from the last student to the end of the room (`n - 1 - students[-1]`).
Step 6: Compare all three distances (start, middle, end) to find the maximum distance. The corresponding seat (either 0, the midpoint, or n-1) is the best seat.
Step 7: Insert the best seat into the `students` list in sorted order using `bisect.insort` (Python) or equivalent methods in Java/C++.
Step 8: In the `leave(p)` method, remove the seat number `p` from the `students` list.

Key Insights

Insight 1: Maintaining a sorted list of occupied seats allows efficient calculation of distances to neighboring students.
Insight 2: The maximum distance can occur at the beginning of the row (seat 0), in the middle between two students, or at the end of the row (seat n-1).
Insight 3: Binary search (or similar sorted insertion) is crucial for efficiently placing the student in the correct sorted position within the 'students' list.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Design, Heap (Priority Queue), Ordered Set.

Companies

Asked at: Quora.

Exam Room | LeetCode 855

Medium

Design Heap (Priority Queue)Ordered Set

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class ExamRoom {

    private int n;
    private List<Integer> students;

    public ExamRoom(int n) {
        this.n = n;
        this.students = new ArrayList<>();
    }
    
    public int seat() {
        if (students.isEmpty()) {
            students.add(0);
            return 0;
        }

        int max_dist = students.get(0);
        int best_seat = 0;

        for (int i = 0; i < students.size() - 1; i++) {
            int dist = (students.get(i+1) - students.get(i)) / 2;
            if (dist > max_dist) {
                max_dist = dist;
                best_seat = students.get(i) + dist;
            }
        }
        
        int dist_end = (n - 1) - students.get(students.size() - 1);
        if (dist_end > max_dist) {
            best_seat = n - 1;
        }

        int index = Collections.binarySearch(students, best_seat);
        if (index < 0) {
            index = -index - 1;
        }
        students.add(index, best_seat);
        
        return best_seat;
    }
    
    public void leave(int p) {
        students.remove(Integer.valueOf(p));
    }
}

/**
 * Your ExamRoom object will be instantiated and called as such:
 * ExamRoom obj = new ExamRoom(n);
 * int param_1 = obj.seat();
 * obj.leave(p);
 */

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Problem Info

Description

Problem 855: Exam Room - Detailed explanation and solution approaches.

Topics

Design Heap (Priority Queue)Ordered Set

Companies

Quora

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:855

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize the ExamRoom with 'n' total seats and an empty list called `students` to track occupied seats.
2
Step 2: In the `seat()` method, if `students` is empty, return seat 0 and add 0 to the `students` list.
3
Step 3: If `students` is not empty, calculate the distance from seat 0 to the first student (`students[0]`).
4
Step 4: Iterate through the `students` list, calculating the distance between each pair of adjacent students (`students[i+1] - students[i]`) and divide by 2 to find the midpoint distance.
5
Step 5: Calculate the distance from the last student to the end of the room (`n - 1 - students[-1]`).
6
Step 6: Compare all three distances (start, middle, end) to find the maximum distance. The corresponding seat (either 0, the midpoint, or n-1) is the best seat.
7
Step 7: Insert the best seat into the `students` list in sorted order using `bisect.insort` (Python) or equivalent methods in Java/C++.
8
Step 8: In the `leave(p)` method, remove the seat number `p` from the `students` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `seat()` method iterates through the `students` list to calculate the maximum distance. The `leave()` method iterates through the `students` list to find and remove a student. In the worst case, both operations require examining all 'n' students, leading to O(n) time complexity. The space complexity is O(n) because the `students` list can grow to hold up to 'n' student seat positions.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

TreeSet/Ordered Set

Using a TreeSet (Java) or an ordered set (C++) could improve the time complexity of insertion and deletion to O(log n). This data structure maintains a sorted set of student positions and allows efficient insertion and deletion.

Time: O(log n) for `seat()` and `leave()`. The `seat()` method needs to find the correct position for the new student, and the `leave()` method needs to remove the student. With a sorted set, these operations can be done in logarithmic time.Space: O(n)

Trade-offs: Using a TreeSet/ordered set provides better time complexity for insertion and deletion, but it may be more complex to implement.

Priority Queue (Heap)

Instead of storing individual seats, store intervals between occupied seats in a priority queue, sorted by the interval length. The 'seat' operation would then find the largest interval, split it into two, and add the new intervals back into the priority queue. The 'leave' operation would require merging adjacent intervals.

Time: O(log n) for `seat()` and `leave()`. Heap operations are generally O(log n).Space: O(n)

Trade-offs: Priority queue approach can be more complex to reason about, requiring careful consideration of interval updates, especially for the boundary intervals.

✨

Key Insights

1Insight 1: Maintaining a sorted list of occupied seats allows efficient calculation of distances to neighboring students.
2Insight 2: The maximum distance can occur at the beginning of the row (seat 0), in the middle between two students, or at the end of the row (seat n-1).
3Insight 3: Binary search (or similar sorted insertion) is crucial for efficiently placing the student in the correct sorted position within the 'students' list.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: Empty exam room. The `seat()` method correctly assigns seat 0 in this case.
Edge case 2: Only one student is seated. The algorithm correctly considers the distances from the student to the beginning and end of the room.
Edge case 3: The best seat is at the beginning or end of the room. The algorithm correctly identifies and returns these seats.
Edge case 4: Integer overflow. When calculating the distance between students, ensure the difference doesn't exceed integer limits. (Not an issue here because of constraints)

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the distances to the beginning and end of the exam room. Many solutions only focus on the distances between existing students.

✗Common mistake 2: Not maintaining the sorted order of the `students` list. Incorrect placement of new students can lead to incorrect distance calculations.

✗Common mistake 3: Not handling the empty room edge case correctly, leading to errors when the first student tries to sit.

✗Common mistake 4: Integer overflow when calculating distances. Although constraints limit this in this problem, it's important to be aware of this potential issue.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in a distributed system. The algorithm can be used to allocate resources to clients in a way that maximizes the distance between clients to minimize interference.

🏦Real-world scenario 2: Seat assignment in a movie theater or airplane. The algorithm can be used to assign seats to customers in a way that maximizes their personal space.

🎮Real-world scenario 3: Placing cell towers to maximize coverage and minimize signal interference between towers.

📱Real-world scenario 4: Server allocation where maximizing distance could represent load balancing or minimizing resource contention.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Exam Room"?

For this medium-level Design problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether heap (priority queue) techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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