Arrays - Design Parking System

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. carType can be of three kinds: big, medium, or small, which are represented by 1, 2, and 3 respectively. A car can only park in a parking space of its carType. If there is no space available, return false, else park the car in that size space and return true.

Input: Operations in this order
"ParkingSystem": [1, 1, 0]
"addCar" : [1]
"addCar" : [2]
"addCar" : [3]
"addCar" : [1]
Output:
Operation 1 "addCar" : true
Operation 2 "addCar" : true
Operation 3 "addCar" : false
Operation 4 "addCar" : false
Explanation:
parkingSystem.addCar(1); // return true because there is 1 available slot for a big car
parkingSystem.addCar(2); // return true because there is 1 available slot for a medium car
parkingSystem.addCar(3); // return false because there is no available slot for a small car
parkingSystem.addCar(1); // return false because there is no available slot for a big car. (available one already occupied)

Constraints:

0 <= big, medium, small <= 1000
carType is 1, 2, or 3
At most 1000 calls will be made to addCar

Solution 1 - Using a temporary array to store initial count of parking slots

In this approach, we will be creating temporary array with size 3 to store the parking slots count for each type of car, i.e. big, medium and small. Then in addCar(int carType) method, we will check whether there are any parking slots available for the carType passed in, if there are slots available then we will decrement the count by 1 and return true, otherwise return false.

public class ParkingSystem { private int[] parkingSlots = new int[3]; public ParkingSystem(int big, int medium, int small) { parkingSlots[0] = big; parkingSlots[1] = medium; parkingSlots[2] = small; } public boolean addCar(int carType) { if(parkingSlots[carType-1] >0) { parkingSlots[carType - 1] -= 1; // reduce count by 1 once the slot occupied return true; } return false; } public static void main(String[] args) { ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0); System.out.println(parkingSystem.addCar(1)); System.out.println(parkingSystem.addCar(2)); System.out.println(parkingSystem.addCar(3)); System.out.println(parkingSystem.addCar(1)); } }

Complexity Analysis:

Time complexity: Above code runs in O(1) time, because there are only 3 types of car parking slots in the problem.
Space complexity: O(1), since we are only creating a temporary array with size 3.

Above implementations source code can be found at GitHub link for Java code

Constraints:

Contents

Complexity Analysis: