Super keyword in Java - Time & Space Complexity

Time Complexity: Super keyword

O(1)

Understanding Time Complexity

Let's explore how using the super keyword affects the time it takes for a program to run.

We want to see how calling a parent class method with super changes the work done as input grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Parent {
    void display() {
        System.out.println("Parent display");
    }
}

class Child extends Parent {
    void display() {
        super.display();
        System.out.println("Child display");
    }
}

public class Test {
    public static void main(String[] args) {
        Child c = new Child();
        c.display();
    }
}
    

This code calls a method in the child class that also calls the parent class method using super.

Identify Repeating Operations

Look for repeated actions that take time.

• Primary operation: Calling display() method once.
• How many times: The method runs once, calling parent method once inside.

How Execution Grows With Input

Since the method calls happen only once, the work stays the same no matter the input size.

Input Size (n)	Approx. Operations
10	2 method calls
100	2 method calls
1000	2 method calls

Pattern observation: The number of operations does not increase with input size.

Final Time Complexity

Time Complexity: O(1)

This means the time taken stays constant no matter how big the input is.

Common Mistake

[X] Wrong: "Using super makes the program slower as input grows."

[OK] Correct: Calling a parent method with super is just one extra step and does not repeat or grow with input size.

Interview Connect

Understanding how super works helps you explain method calls clearly and shows you know how inheritance affects program flow.

Self-Check

"What if the display() method called super.display() inside a loop that runs n times? How would the time complexity change?"

Practice

1. What does the super keyword do in Java?

easy

A. It defines a static method in the class.

B. It creates a new object of the child class.

C. It accesses methods and variables from the parent class.

D. It terminates the program execution.

Solution

1. Step 1: Understand the role of super

   The super keyword is used to refer to the parent class's members (methods or variables) from a child class.

2. Step 2: Compare options with definition

   Only It accesses methods and variables from the parent class. correctly describes this behavior. Other options describe unrelated actions.

3. Final Answer:

   It accesses methods and variables from the parent class. -> Option C

4. Quick Check:

   super accesses parent members = A [OK]

Hint: Remember: super means parent class access [OK]

Common Mistakes:

• Thinking super creates new objects
• Confusing super with this keyword
• Assuming super ends program

2. Which of the following is the correct way to call a parent class constructor in Java?

easy

A. super();

B. this();

C. parent();

D. base();

Solution

1. Step 1: Recall syntax for parent constructor call

   In Java, super() is used inside a child constructor to call the parent class constructor.

2. Step 2: Evaluate options

   Only super(); uses the correct keyword super(). Others are invalid or refer to different concepts.

3. Final Answer:

   super(); -> Option A

4. Quick Check:

   Parent constructor call = super() [OK]

Hint: Use super() to call parent constructor [OK]

Common Mistakes:

• Using this() instead of super()
• Trying to call parent() which is invalid
• Confusing base() with super()

3. What will be the output of the following code?

class Parent {
  int x = 10;
}
class Child extends Parent {
  int x = 20;
  void printX() {
    System.out.println(super.x);
  }
}
public class Test {
  public static void main(String[] args) {
    Child c = new Child();
    c.printX();
  }
}

medium

A. 20

B. Compilation error

C. 0

D. 10

Solution

1. Step 1: Understand variable hiding and super usage

   The child class has its own x = 20, but super.x accesses the parent's x which is 10.

2. Step 2: Trace the print statement

   The method printX() prints super.x, so it prints 10.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   super.x accesses parent variable = 10 [OK]

Hint: super.variable accesses parent class variable [OK]

Common Mistakes:

• Printing child variable instead of parent
• Confusing super.x with this.x
• Expecting compilation error

4. Identify the error in this code snippet:

class Parent {
  void show() {
    System.out.println("Parent show");
  }
}
class Child extends Parent {
  void show() {
    super.show();
    System.out.println("Child show");
  }
  void display() {
    super();
  }
}

medium

A. super.show() is invalid inside child class

B. super() cannot be called like a method in display()

C. Child class cannot override show() method

D. No error, code is correct

Solution

1. Step 1: Check usage of super in methods

   Calling super.show() inside overridden method is valid to call parent method.

2. Step 2: Analyze super() call in display()

   super() can only be used to call parent constructor inside child constructor, not as a method call elsewhere.

3. Final Answer:

   super() cannot be called like a method in display() -> Option B

4. Quick Check:

   super() only in constructor = D [OK]

Hint: super() only calls parent constructor inside child constructor [OK]

Common Mistakes:

• Using super() outside constructor
• Thinking super.show() is invalid
• Believing overriding is not allowed

5. Given these classes:

class Animal {
  String name;
  Animal(String name) {
    this.name = name;
  }
  void sound() {
    System.out.println("Animal sound");
  }
}
class Dog extends Animal {
  Dog() {
    super("Dog");
  }
  void sound() {
    super.sound();
    System.out.println("Bark");
  }
}
public class Test {
  public static void main(String[] args) {
    Dog d = new Dog();
    d.sound();
    System.out.println(d.name);
  }
}

What is the output when running Test.main()?

hard

A. Animal sound
Bark
Dog

B. Bark
Animal sound
Dog

C. Animal sound
Dog
Bark

D. Compilation error due to super()

Solution

1. Step 1: Understand constructor chaining

   The Dog constructor calls super("Dog"), setting name to "Dog" in Animal.

2. Step 2: Trace the sound() method call

   Dog.sound() calls super.sound() which prints "Animal sound", then prints "Bark".

3. Step 3: Print the name field

   Printing d.name outputs "Dog".

4. Final Answer:

   Animal sound Bark Dog -> Option A

5. Quick Check:

   super() sets name, super.sound() prints parent sound = A [OK]

Hint: super() sets parent state; super.method() calls parent method [OK]

Common Mistakes:

• Expecting Dog before Animal sound
• Confusing order of prints
• Thinking super() causes error