LeetCode #1603 — EASY

Design Parking System

Build confidence with an intuition-first walkthrough focused on design fundamentals.

The Problem

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. 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.

Example 1:

Input
["ParkingSystem", "addCar", "addCar", "addCar", "addCar"]
[[1, 1, 0], [1], [2], [3], [1]]
Output
[null, true, true, false, false]

Explanation
ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0);
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. It is already occupied.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Design

Example 1

["ParkingSystem","addCar","addCar","addCar","addCar"]
[[1,1,0],[1],[2],[3],[1]]

Step 02

Core Insight

What unlocks the optimal approach

Record number of parking slots still available for each car type.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1603: Design Parking System
class ParkingSystem {
    private int[] cnt;

    public ParkingSystem(int big, int medium, int small) {
        cnt = new int[] {0, big, medium, small};
    }

    public boolean addCar(int carType) {
        if (cnt[carType] == 0) {
            return false;
        }
        --cnt[carType];
        return true;
    }
}

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

// Accepted solution for LeetCode #1603: Design Parking System
type ParkingSystem struct {
	cnt []int
}

func Constructor(big int, medium int, small int) ParkingSystem {
	return ParkingSystem{[]int{0, big, medium, small}}
}

func (this *ParkingSystem) AddCar(carType int) bool {
	if this.cnt[carType] == 0 {
		return false
	}
	this.cnt[carType]--
	return true
}

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

# Accepted solution for LeetCode #1603: Design Parking System
class ParkingSystem:
    def __init__(self, big: int, medium: int, small: int):
        self.cnt = [0, big, medium, small]

    def addCar(self, carType: int) -> bool:
        if self.cnt[carType] == 0:
            return False
        self.cnt[carType] -= 1
        return True


# Your ParkingSystem object will be instantiated and called as such:
# obj = ParkingSystem(big, medium, small)
# param_1 = obj.addCar(carType)

// Accepted solution for LeetCode #1603: Design Parking System
struct ParkingSystem {
    cnt: [i32; 4],
}

impl ParkingSystem {
    fn new(big: i32, medium: i32, small: i32) -> Self {
        ParkingSystem {
            cnt: [0, big, medium, small],
        }
    }

    fn add_car(&mut self, car_type: i32) -> bool {
        if self.cnt[car_type as usize] == 0 {
            return false;
        }
        self.cnt[car_type as usize] -= 1;
        true
    }
}

// Accepted solution for LeetCode #1603: Design Parking System
class ParkingSystem {
    private cnt: [number, number, number, number];

    constructor(big: number, medium: number, small: number) {
        this.cnt = [0, big, medium, small];
    }

    addCar(carType: number): boolean {
        if (this.cnt[carType] === 0) {
            return false;
        }
        this.cnt[carType]--;
        return true;
    }
}

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

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.