How do you solve Design Parking System on LeetCode?

Design Parking System (LeetCode #1603) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Using an array to track available slots is efficient.

What is the time complexity of Design Parking System?

The optimal solution for Design Parking System runs in O(1) time and uses O(1) space. Both time and space complexity are O(1) because we are only using a fixed-size array to store the counts of the parking slots.

What is the brute force solution for Design Parking System?

The brute force approach for Design Parking System is Brute Force, with O(1) time complexity and O(1) space complexity. The brute force approach involves checking each parking space type sequentially to see if there's an available slot for the car type. This is straightforward but inefficient as it checks all conditions every time a car tries to park. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Design Parking System?

Design Parking System uses the following concepts: Design, Simulation, Counting. The recommended patterns to study are: Array, Simulation.

What are the interview tips for Design Parking System?

Explain your thought process clearly. Consider edge cases, like when all slots are full. Be ready to discuss time and space complexity.

What are common mistakes when solving Design Parking System?

Not initializing the slots correctly. Confusing the carType index when accessing the slots.

#1603

Design Parking System

Easy

DesignSimulationCountingArraySimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(1)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

The optimal solution utilizes a simple array to track available slots for each car type. This allows for constant time checks and updates, making it efficient for multiple parking requests.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array with the number of available slots for big, medium, and small cars.
2Step 2: For each addCar call, check the corresponding index in the array.
3Step 3: If a slot is available, decrement the count and return true; otherwise, return false.

solution.py9 lines

1class ParkingSystem:
2    def __init__(self, big: int, medium: int, small: int):
3        self.slots = [big, medium, small]
4
5    def addCar(self, carType: int) -> bool:
6        if self.slots[carType - 1] > 0:
7            self.slots[carType - 1] -= 1
8            return True
9        return False

ℹ

Complexity note: Both time and space complexity are O(1) because we are only using a fixed-size array to store the counts of the parking slots.

1Using an array to track available slots is efficient.
2Direct indexing allows for constant time operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.