What is the time complexity of Design Parking System?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Design Parking System?

This problem uses Design, Simulation, Counting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Design Parking System?

Insight 1: Using an array to store the available slots for each car type simplifies the code and makes access to the relevant slot count O(1). Insight 2: The problem's constraints (limited number of car types and calls to `addCar`) allow for a simple and efficient solution without needing more complex data structures.

Which companies ask LeetCode 1603. Design Parking System in interviews?

This problem is frequently asked at Valve.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Design Parking System - LeetCode 1603 Solution

Design Parking System - Complete Solution Guide

Design Parking System is LeetCode problem 1603, a Easy 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 parking system for a parking lot. The parking lot has three kinds of parking spaces: big, medium, and small, with a fixed number of slots for each size. Implement the ParkingSystem class: ParkingSystem(int big, int medium, int small) Initializes object of the ParkingSystem class. The number of slots for each parking space are given as part of the constructor. bool addCar(int carType) Checks whether there is a parking space of carType for the car that wants to get into the parking lot. c

Detailed Explanation

The LeetCode problem 1603, "Design Parking System," asks you to design a system that manages parking spaces in a lot with three types of spaces: big, medium, and small. The system should initialize with a given number of slots for each type and provide a method to check if a car of a specific type can park and update the available slots accordingly. The input consists of the initial number of slots for each type and a series of car types requesting parking (1 for big, 2 for medium, 3 for small). The output is a boolean indicating whether parking was successful for each car.

Solution Approach

The provided solutions use a straightforward approach. They initialize an array (or individual variables in Java and C++) to store the number of available slots for each car type (big, medium, small). The `addCar` method then checks if there is an available slot of the requested `carType` by looking up the corresponding index in the array. If a slot is available, it decrements the counter for that slot type and returns `true`; otherwise, it returns `false`. This direct approach ensures constant-time operations for both initialization and car addition.

Step-by-Step Algorithm

Step 1: Initialize the ParkingSystem object with the number of big, medium, and small slots. Store these values in an array (Python) or individual variables (Java, C++, C).
Step 2: When the `addCar` method is called with a `carType`, check if the corresponding slot count (e.g., `slots[carType - 1]` in Python) is greater than 0.
Step 3: If a slot is available, decrement the count and return `true` indicating successful parking.
Step 4: If no slot is available, return `false`.

Key Insights

Insight 1: Using an array to store the available slots for each car type simplifies the code and makes access to the relevant slot count O(1).
Insight 2: The problem's constraints (limited number of car types and calls to `addCar`) allow for a simple and efficient solution without needing more complex data structures.
Insight 3: Handling the edge case where the requested car type has no available slots is crucial for correct functionality. Returning `false` in such scenarios is essential.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Design, Simulation, Counting.

Companies

Asked at: Valve.

Design Parking System | LeetCode 1603

Easy

Design Simulation Counting

Time: O(1)Space: O(1)

Solution Code

Available in:

class ParkingSystem {

    private int big;
    private int medium;
    private int small;

    public ParkingSystem(int big, int medium, int small) {
        this.big = big;
        this.medium = medium;
        this.small = small;
    }
    
    public boolean addCar(int carType) {
        if (carType == 1) {
            if (big > 0) {
                big--;
                return true;
            } else {
                return false;
            }
        } else if (carType == 2) {
            if (medium > 0) {
                medium--;
                return true;
            } else {
                return false;
            }
        } else {
            if (small > 0) {
                small--;
                return true;
            } else {
                return false;
            }
        }
    }
}

/**
 * Your ParkingSystem object will be instantiated and called as such:
 * ParkingSystem obj = new ParkingSystem(big, medium, small);
 * boolean param_1 = obj.addCar(carType);
 */

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1603: Design Parking System - Detailed explanation and solution approaches.

Topics

Design Simulation Counting

Companies

Valve

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1603

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

4 steps

1
Step 1: Initialize the ParkingSystem object with the number of big, medium, and small slots. Store these values in an array (Python) or individual variables (Java, C++, C).
2
Step 2: When the `addCar` method is called with a `carType`, check if the corresponding slot count (e.g., `slots[carType - 1]` in Python) is greater than 0.
3
Step 3: If a slot is available, decrement the count and return `true` indicating successful parking.
4
Step 4: If no slot is available, return `false`.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Both the initialization and the `addCar` method involve a constant number of operations regardless of the input size. The array (or variables) used for storing the slot counts occupies a fixed amount of memory.

📖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

Using a HashMap (or similar)

Instead of an array, a hashmap could be used to store the available slots, with the car type as the key and the number of available slots as the value. This approach would be more flexible if the number of car types could change.

Time: O(1) on average, O(n) in worst case (hash collision)Space: O(n), where n is the number of car types

Trade-offs: While offering more flexibility, a HashMap adds overhead compared to the direct array access used in the provided solution. The array approach is superior for this problem's constraints.

✨

Key Insights

1Insight 1: Using an array to store the available slots for each car type simplifies the code and makes access to the relevant slot count O(1).
2Insight 2: The problem's constraints (limited number of car types and calls to `addCar`) allow for a simple and efficient solution without needing more complex data structures.
3Insight 3: Handling the edge case where the requested car type has no available slots is crucial for correct functionality. Returning `false` in such scenarios is essential.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: All slots of a particular type are occupied. The solution correctly returns `false` when `addCar` is called for a car type with zero available slots.
Edge case 2: Invalid `carType` (not 1, 2, or 3). The C solution explicitly handles this, while others implicitly handle it due to array indexing. Explicitly checking for valid car types adds robustness.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where there are no available slots of the requested type. The solution must explicitly check for this condition and return `false`.

✗Common mistake 2: Incorrect array indexing. In the array-based solution, remember to adjust the index using `carType - 1` to access the correct slot count, considering that arrays are 0-indexed.

✅

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: Managing parking spaces in a real-world parking lot. This system can be a simplified version of a larger parking management system.

🏦Real-world scenario 2: Resource allocation in various systems. This problem's concept of managing a limited number of resources and assigning them based on type can apply to other scenarios, like assigning tasks to workers based on their skills.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

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

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: November 11, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Design Parking System"?

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

🤖 Need help with this problem?