Base class and subclass in Swift - Time & Space Complexity

Time Complexity: Base class and subclass

O(n)

Understanding Time Complexity

When using base classes and subclasses, it's important to see how the program's work grows as input changes.

We want to know how the time to run the code changes when we create or use many objects.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Animal {
    func speak() {
        print("Animal sound")
    }
}

class Dog: Animal {
    override func speak() {
        print("Bark")
    }
}

let dogs = (1...n).map { _ in Dog() }
dogs.forEach { $0.speak() }
    

This code creates n Dog objects and calls their speak method one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Creating n Dog objects and calling speak on each.
• How many times: The speak method is called once for each Dog, so n times.

How Execution Grows With Input

As the number of Dog objects increases, the total work grows in a straight line.

Input Size (n)	Approx. Operations
10	About 10 speak calls
100	About 100 speak calls
1000	About 1000 speak calls

Pattern observation: Doubling n doubles the work because each object is handled once.

Final Time Complexity

Time Complexity: O(n)

This means the time grows directly with the number of objects created and used.

Common Mistake

[X] Wrong: "Because of inheritance, the code runs slower exponentially as n grows."

[OK] Correct: Inheritance itself does not add extra loops or repeated work; each object is still handled once.

Interview Connect

Understanding how subclassing affects time helps you explain object-oriented code performance clearly and confidently.

Self-Check

"What if the speak method called another method that loops over all dogs? How would the time complexity change?"

Practice

1. What is the main purpose of a base class in Swift?

easy

A. To prevent other classes from inheriting

B. To provide a general template for other classes to build upon

C. To store data only without any functions

D. To create instances that cannot be used

Solution

1. Step 1: Understand the role of a base class

   A base class defines common properties and methods that other classes can reuse.

2. Step 2: Compare options with this role

   To provide a general template for other classes to build upon correctly states the base class is a general template. Other options describe incorrect or unrelated roles.

3. Final Answer:

   To provide a general template for other classes to build upon -> Option B

4. Quick Check:

   Base class = general template [OK]

Hint: Base class = general template for subclasses [OK]

Common Mistakes:

• Thinking base class cannot be inherited
• Confusing base class with instances
• Assuming base class only stores data

2. Which of the following is the correct way to declare a subclass named Dog inheriting from a base class Animal in Swift?

easy

A. class Dog inherits Animal { }

B. class Dog extends Animal { }

C. class Dog: Animal { }

D. class Dog < Animal { }

Solution

1. Step 1: Recall Swift subclass syntax

   In Swift, a subclass is declared using a colon (:) after the subclass name followed by the base class name.

2. Step 2: Match options to syntax

   class Dog: Animal { } uses class Dog: Animal { }, which is correct. Other options use incorrect keywords or symbols.

3. Final Answer:

   class Dog: Animal { } -> Option C

4. Quick Check:

   Subclass syntax = colon (:) [OK]

Hint: Use colon (:) to inherit in Swift class declaration [OK]

Common Mistakes:

• Using 'inherits' or 'extends' keywords (not Swift)
• Using wrong symbols like <
• Omitting the colon between subclass and base class

3. Consider this Swift code:

class Vehicle {
    func sound() -> String {
        return "Vroom"
    }
}

class Car: Vehicle {
    override func sound() -> String {
        return "Beep"
    }
}

let myCar = Car()
print(myCar.sound())

What will be printed when this code runs?

medium

A. "Beep"

B. "Vroom"

C. Compilation error

D. "Car"

Solution

1. Step 1: Understand method overriding in subclass

   The subclass Car overrides the sound() method to return "Beep" instead of the base class "Vroom".

2. Step 2: Identify which method is called

   When calling myCar.sound(), the overridden method in Car runs, returning "Beep".

3. Final Answer:

   "Beep" -> Option A

4. Quick Check:

   Overridden method runs = "Beep" [OK]

Hint: Overridden methods in subclass replace base methods [OK]

Common Mistakes:

• Expecting base class method output
• Confusing method override with overload
• Thinking print shows class name

4. Identify the error in this Swift subclass declaration:

class Animal {
    func speak() {
        print("Animal sound")
    }
}

class Cat: Animal {
    func speak() {
        print("Meow")
    }
}

medium

A. Class names must be lowercase

B. Base class cannot have methods

C. Subclass cannot have methods with same name

D. Missing override keyword before speak() in subclass

Solution

1. Step 1: Check method overriding rules in Swift

   When a subclass provides its own version of a base class method, it must use the override keyword.

2. Step 2: Identify missing override keyword

   The speak() method in Cat lacks override, causing a compile error.

3. Final Answer:

   Missing override keyword before speak() in subclass -> Option D

4. Quick Check:

   Override keyword required for method override [OK]

Hint: Use override keyword when redefining base methods [OK]

Common Mistakes:

• Forgetting override keyword
• Thinking subclass methods cannot have same name
• Assuming class names must be lowercase

5. You want to create a base class Employee with a property name and a method work() that prints "Working". Then create a subclass Manager that overrides work() to print "Managing team" and adds a new method holdMeeting() that prints "Meeting started".

Which code correctly implements this?

hard

A. class Employee {
var name: String
init(name: String) {
self.name = name
}
func work() {
print("Working")
}
}

class Manager: Employee {
override func work() {
print("Managing team")
}
func holdMeeting() {
print("Meeting started")
}
}

B. class Employee {
var name: String
func work() {
print("Working")
}
}

class Manager: Employee {
func work() {
print("Managing team")
}
func holdMeeting() {
print("Meeting started")
}
}

C. class Employee {
var name: String
init(name: String) {
self.name = name
}
func work() {
print("Working")
}
}

class Manager: Employee {
func work() {
print("Managing team")
}
func holdMeeting() {
print("Meeting started")
}
}

D. class Employee {
var name: String
init(name: String) {
self.name = name
}
func work() {
print("Working")
}
}

class Manager: Employee {
override func work() {
print("Managing team")
}
}

Solution

1. Step 1: Check base class implementation

   The base class Employee must have a name property with initializer and a work() method printing "Working". class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } and C have correct base class with initializer.

2. Step 2: Check subclass overrides and new method

   The subclass Manager must override work() with override keyword and add holdMeeting(). class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } correctly uses override and adds holdMeeting(). class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } misses override. class Employee { var name: String func work() { print("Working") } } class Manager: Employee { func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } misses initializer in base class. class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } } misses holdMeeting().

3. Final Answer:

   Option A code correctly implements all requirements -> Option A

4. Quick Check:

   Override + new method + initializer = class Employee { var name: String init(name: String) { self.name = name } func work() { print("Working") } } class Manager: Employee { override func work() { print("Managing team") } func holdMeeting() { print("Meeting started") } } [OK]

Hint: Use override keyword and add new methods in subclass [OK]

Common Mistakes:

• Forgetting override keyword in subclass
• Missing initializer in base class
• Not adding new methods in subclass