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Default access modifier in Java - Time & Space Complexity

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Time Complexity: Default access modifier
O(1)
Understanding Time Complexity

Let's see how the default access modifier affects the time complexity of code access.

We want to know how the cost of accessing members changes with input size.

Scenario Under Consideration

Analyze the time complexity of accessing a default access modifier member in a class.

class Box {
    int width;  // default access

    int getWidth() {
        return width;
    }
}

public class Main {
    public static void main(String[] args) {
        Box box = new Box();
        box.width = 10;
        System.out.println(box.getWidth());
    }
}

This code defines a class with a default access field and accesses it.

Identify Repeating Operations

Look for repeated actions that affect time.

  • Primary operation: Accessing the width field and calling getWidth().
  • How many times: Once in this example, but could be many times if in a loop.
How Execution Grows With Input

Accessing a field or method does not depend on input size.

Input Size (n)Approx. Operations
1010 accesses if repeated 10 times
100100 accesses if repeated 100 times
10001000 accesses if repeated 1000 times

Pattern observation: Each access is constant time, so total time grows linearly with number of accesses.

Final Time Complexity

Time Complexity: O(1)

This means accessing a default access field or method takes the same time no matter the input size.

Common Mistake

[X] Wrong: "Accessing a default access field takes longer as the program grows."

[OK] Correct: Access time depends on the operation itself, not on how many other classes or files exist.

Interview Connect

Understanding access modifiers helps you reason about code organization and performance clearly, a useful skill in many coding tasks.

Self-Check

What if the width field was declared private and accessed via a getter method? How would the time complexity change?

Practice

(1/5)
1. Which statement best describes the default access modifier in Java?
easy
A. It restricts access only to the class itself.
B. It allows access from any package using the default keyword.
C. It allows access only within the same package without using any keyword.
D. It allows access only to subclasses in other packages.

Solution

  1. Step 1: Understand default access modifier meaning

    The default access modifier means no keyword is used before a class, method, or variable, and it restricts access to the same package only.
  2. Step 2: Compare options with definition

    It allows access only within the same package without using any keyword. correctly states it allows access only within the same package and requires no keyword. Other options mention keywords or access outside the package, which is incorrect.
  3. Final Answer:

    It allows access only within the same package without using any keyword. -> Option C
  4. Quick Check:

    Default access = package-only access [OK]
Hint: No keyword means default access is package-only [OK]
Common Mistakes:
  • Thinking default requires a keyword
  • Confusing default with public or private
  • Assuming default allows access from other packages
2. Which of the following is the correct way to declare a class with default access modifier?
easy
A. class MyClass {}
B. private class MyClass {}
C. default class MyClass {}
D. public class MyClass {}

Solution

  1. Step 1: Recall syntax for default access modifier

    Default access means no access keyword is written before the class name.
  2. Step 2: Check each option

    class MyClass {} declares the class without any access keyword, which means default access. Options A and B use explicit keywords, and C uses a non-existent keyword.
  3. Final Answer:

    class MyClass {} -> Option A
  4. Quick Check:

    No keyword = default access [OK]
Hint: No keyword before class means default access [OK]
Common Mistakes:
  • Using 'default' keyword which does not exist in Java
  • Confusing private with default
  • Always using public by habit
3. Consider two classes in the same package:
class A {
    void show() {
        System.out.println("Hello from A");
    }
}

class B {
    public static void main(String[] args) {
        A a = new A();
        a.show();
    }
}

What will be the output when class B is run?
medium
A. Compile-time error: show() has private access
B. Hello from A
C. Compile-time error: class A is not public
D. Runtime error

Solution

  1. Step 1: Analyze access modifiers in the code

    Class A and method show() have default access (no keyword). Both classes are in the same package, so access is allowed.
  2. Step 2: Predict program output

    Since B can access A and its show() method, the program prints "Hello from A" without errors.
  3. Final Answer:

    Hello from A -> Option B
  4. Quick Check:

    Default access allows same package access [OK]
Hint: Default access works within same package [OK]
Common Mistakes:
  • Assuming default means private access
  • Thinking class A must be public to be used
  • Confusing compile-time and runtime errors
4. Given two packages pkg1 and pkg2, and these classes:
package pkg1;
class C {
    void display() {
        System.out.println("Inside C");
    }
}

package pkg2;
import pkg1.C;
public class D {
    public static void main(String[] args) {
        C c = new C();
        c.display();
    }
}

What is the error in this code?
medium
A. Cannot access class C from pkg2 because it has default access
B. Cannot call display() because it is private
C. Cannot import package pkg1
D. No error, code runs fine

Solution

  1. Step 1: Check class C's access modifier and package

    Class C has default access (no keyword) and is in package pkg1, so it is accessible only within pkg1.
  2. Step 2: Check usage of C in pkg2

    Class D in pkg2 tries to create an instance of C and call display(), but C is not visible outside pkg1, causing a compile-time error.
  3. Final Answer:

    Cannot access class C from pkg2 because it has default access -> Option A
  4. Quick Check:

    Default access = package-only visibility [OK]
Hint: Default access blocks cross-package access [OK]
Common Mistakes:
  • Assuming default access allows cross-package use
  • Thinking display() is private
  • Confusing import errors with access errors
5. You have a package tools with classes:
class Helper {
    void assist() {
        System.out.println("Helping...");
    }
}

public class Tool {
    public static void main(String[] args) {
        Helper h = new Helper();
        h.assist();
    }
}

You want to allow only classes inside tools to use Helper and its assist() method, but prevent access from outside. Which is the best way to achieve this?
hard
A. Make Helper and assist() public.
B. Use protected for Helper and assist().
C. Make Helper and assist() private.
D. Keep Helper and assist() with default access (no keyword).

Solution

  1. Step 1: Understand access goals

    You want Helper and assist() accessible only inside the same package (tools) and hidden outside.
  2. Step 2: Match access modifiers to goals

    Default access (no keyword) restricts access to the same package only, which fits the requirement. Public allows all access, private restricts to the class only, and protected allows subclasses and package access.
  3. Final Answer:

    Keep Helper and assist() with default access (no keyword). -> Option D
  4. Quick Check:

    Default access = package-only visibility [OK]
Hint: Default access hides from outside packages [OK]
Common Mistakes:
  • Using public which exposes to all packages
  • Using private which hides even inside package
  • Using protected which allows subclass access outside package
Recommended

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10 questions~5 mineasy hard