What is the time complexity of Seat Reservation Manager?

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

What data structures are used to solve Seat Reservation Manager?

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

What are the key insights for solving Seat Reservation Manager?

Insight 1: The smallest-numbered available seat is crucial, suggesting the use of a data structure that can efficiently find the minimum element. Insight 2: A min-heap (priority queue) is a suitable data structure to maintain the set of available seats, as it provides O(log n) time complexity for insertion (unreserve) and extracting the minimum element (reserve).

Which companies ask LeetCode 1845. Seat Reservation Manager in interviews?

This problem is frequently asked at Dropbox, Siemens.

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Seat Reservation Manager - LeetCode 1845 Solution

Seat Reservation Manager - Complete Solution Guide

Seat Reservation Manager is LeetCode problem 1845, 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

Design a system that manages the reservation state of n seats that are numbered from 1 to n . Implement the SeatManager class: SeatManager(int n) Initializes a SeatManager object that will manage n seats numbered from 1 to n . All seats are initially available. int reserve() Fetches the smallest-numbered unreserved seat, reserves it, and returns its number. void unreserve(int seatNumber) Unreserves the seat with the given seatNumber . Example 1: Input ["SeatManager", "reserve", "reserve", "unres

Detailed Explanation

The problem asks us to design a system, the `SeatManager`, which manages the reservation status of `n` seats, numbered from 1 to `n`. Initially, all seats are available. The `SeatManager` needs to implement two core functionalities: `reserve()` and `unreserve(seatNumber)`. The `reserve()` function should return the smallest-numbered available seat, reserving it in the process. The `unreserve(seatNumber)` function should make the specified seat available again.

Solution Approach

The provided code uses a min-heap (priority queue) to store the available seats. When a seat is reserved, if there are available seats in the heap, the smallest seat number is extracted from the heap and returned. Otherwise, the next available seat number is used, and the `next_seat` counter is incremented. When a seat is unreserved, its number is added to the heap. The min-heap property ensures that the `reserve()` function always returns the smallest available seat.

Step-by-Step Algorithm

Step 1: Initialize the `SeatManager` with `n` (number of seats), `next_seat` (starting at 1), and an empty min-heap `available_seats`.
Step 2: In the `reserve()` function, check if the `available_seats` heap is empty.
Step 3: If the heap is not empty, extract the minimum element (smallest available seat) from the heap and return it. This seat is now reserved.
Step 4: If the heap is empty, return the value of `next_seat` and increment `next_seat`. This allocates the next available seat number.
Step 5: In the `unreserve(seatNumber)` function, add the `seatNumber` back into the `available_seats` min-heap. This makes the seat available for reservation again.

Key Insights

Insight 1: The smallest-numbered available seat is crucial, suggesting the use of a data structure that can efficiently find the minimum element.
Insight 2: A min-heap (priority queue) is a suitable data structure to maintain the set of available seats, as it provides O(log n) time complexity for insertion (unreserve) and extracting the minimum element (reserve).
Insight 3: It is important to track the next available seat number to use when the heap is empty. This ensures that seat numbers are assigned in increasing order when all previously unreserved seats have been reserved.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Dropbox, Siemens.

Seat Reservation Manager | LeetCode 1845

Medium

Design Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.PriorityQueue;

class SeatManager {

    private int next_seat;
    private PriorityQueue<Integer> available_seats;

    public SeatManager(int n) {
        this.next_seat = 1;
        this.available_seats = new PriorityQueue<>();
    }
    
    public int reserve() {
        if (!available_seats.isEmpty()) {
            return available_seats.poll();
        }
        
        int seat = next_seat;
        next_seat += 1;
        return seat;
    }
    
    public void unreserve(int seatNumber) {
        available_seats.offer(seatNumber);
    }
}

/**
 * Your SeatManager object will be instantiated and called as such:
 * SeatManager obj = new SeatManager(n);
 * int param_1 = obj.reserve();
 * obj.unreserve(seatNumber);
 */

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Problem Info

Description

Problem 1845: Seat Reservation Manager - Detailed explanation and solution approaches.

Topics

Design Heap (Priority Queue)

Companies

DropboxSiemens

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1845

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

5 steps

1
Step 1: Initialize the `SeatManager` with `n` (number of seats), `next_seat` (starting at 1), and an empty min-heap `available_seats`.
2
Step 2: In the `reserve()` function, check if the `available_seats` heap is empty.
3
Step 3: If the heap is not empty, extract the minimum element (smallest available seat) from the heap and return it. This seat is now reserved.
4
Step 4: If the heap is empty, return the value of `next_seat` and increment `next_seat`. This allocates the next available seat number.
5
Step 5: In the `unreserve(seatNumber)` function, add the `seatNumber` back into the `available_seats` min-heap. This makes the seat available for reservation again.

💡

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(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the heap operations, which have a logarithmic time complexity with respect to the number of elements in the heap. The space complexity is determined by the potential size of the heap, which can grow up to `n` in the worst case where all seats are unreserved.

📖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

Sorted Set

Instead of a min-heap, a sorted set (e.g., `TreeSet` in Java, `set` in C++) could be used to store the available seats. The `reserve()` function would take the first element, and the `unreserve()` function would insert an element while maintaining the sorted order.

Time: O(log n) for reserve() and unreserve()Space: O(n)

Trade-offs: Using a sorted set provides similar performance characteristics to a min-heap. The choice between the two often depends on personal preference and the specific language's implementation details. In some cases, sorted sets might have slightly higher overhead due to the tree structure.

Boolean Array

A boolean array could be used to mark each seat as reserved or unreserved. The `reserve()` function would iterate through the array to find the smallest available seat, and `unreserve()` would set the corresponding element in the array to false.

Time: O(n) for reserve() in the worst case, O(1) for unreserve()Space: O(n)

Trade-offs: The array approach provides O(1) time for `unreserve()` but O(n) time for `reserve()` in the worst case when the first available seat is near the end of the array. This approach is less efficient than using a min-heap or sorted set, especially when there are many `reserve` operations.

✨

Key Insights

1Insight 1: The smallest-numbered available seat is crucial, suggesting the use of a data structure that can efficiently find the minimum element.
2Insight 2: A min-heap (priority queue) is a suitable data structure to maintain the set of available seats, as it provides O(log n) time complexity for insertion (unreserve) and extracting the minimum element (reserve).
3Insight 3: It is important to track the next available seat number to use when the heap is empty. This ensures that seat numbers are assigned in increasing order when all previously unreserved seats have been reserved.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The `available_seats` heap is empty. The solution correctly handles this by assigning the next available seat number and incrementing `next_seat`.
Edge case 2: The initial value of `n` is small (e.g., 1). The solution works correctly, as the heap is initialized with a size appropriate for `n` and the logic for `reserve` and `unreserve` remains valid.
Edge case 3: The `unreserve` operation is called multiple times on the same seat. The heap correctly stores multiple instances of the same seat number, ensuring it can be reserved multiple times.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a min-heap or sorted set. Using a simple list and iterating to find the minimum available seat results in O(n) time complexity for the `reserve()` operation, making the overall solution less efficient.

✗Common mistake 2: Incorrectly implementing the heap operations, particularly `minHeapify`. A faulty heap implementation can lead to incorrect results and significantly impact performance.

✗Common mistake 3: Not handling the case where the heap is empty in the `reserve()` function. This can lead to incorrect seat assignments and violates the problem constraints.

✗Common mistake 4: Failing to free dynamically allocated memory in C. Memory leaks can occur if allocated memory is not properly deallocated.

✅

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 cloud computing environment. Seats can represent virtual machines or containers, and the `SeatManager` can be used to efficiently allocate and deallocate these resources.

🏦Real-world scenario 2: Ticket reservation systems. Seats can represent available seats in a theater or stadium, and the `SeatManager` can be used to manage ticket reservations and cancellations.

🎮Real-world scenario 3: Memory management in operating systems. Seats can represent blocks of memory, and the SeatManager can be used to allocate and deallocate memory blocks efficiently, minimizing fragmentation. This is especially relevant when dealing with variable-sized blocks of memory.

💡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: July 16, 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 "Seat Reservation Manager"?

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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