Using Scanner class in Java - Time & Space Complexity

Time Complexity: Using Scanner class

O(n)

Understanding Time Complexity

When we use the Scanner class to read input, it is important to understand how the time taken grows as the input size increases.

We want to know how the program's running time changes when we read more data.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import java.util.Scanner;

public class InputExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        for (int i = 0; i < n; i++) {
            int value = scanner.nextInt();
            System.out.println(value);
        }
        scanner.close();
    }
}
    

This code reads an integer n, then reads n more integers and prints each one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Reading an integer inside the loop using scanner.nextInt()
• How many times: Exactly n times, where n is the first input read

How Execution Grows With Input

As the number n grows, the number of times we read input and print also grows linearly.

Input Size (n)	Approx. Operations
10	About 10 reads and prints
100	About 100 reads and prints
1000	About 1000 reads and prints

Pattern observation: The work grows directly in proportion to n; doubling n doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows in a straight line with the number of inputs read.

Common Mistake

[X] Wrong: "Reading input with Scanner is always constant time regardless of input size."

[OK] Correct: Each call to scanner.nextInt() reads one integer, so more inputs mean more calls and more time.

Interview Connect

Understanding how input reading scales helps you write efficient programs and shows you can think about performance in real situations.

Self-Check

"What if we changed the code to read all inputs into an array first, then process them? How would the time complexity change?"

Practice

1. What is the primary purpose of the Scanner class in Java?

easy

A. To write output to the console

B. To handle file operations

C. To create graphical user interfaces

D. To read input from the user

Solution

1. Step 1: Understand Scanner's role

   The Scanner class is designed to read input from various sources, commonly from the user via the keyboard.

2. Step 2: Differentiate from other classes

   Writing output is done by System.out, GUI creation uses other libraries, and file handling uses classes like FileReader.

3. Final Answer:

   To read input from the user -> Option D

4. Quick Check:

   Scanner reads input = B [OK]

Hint: Scanner reads user input from keyboard easily [OK]

Common Mistakes:

• Confusing Scanner with output classes
• Thinking Scanner creates GUI
• Assuming Scanner handles files only

2. Which of the following is the correct way to create a Scanner object to read from the keyboard?

easy

A. Scanner input = new Scanner(System.in);

B. Scanner input = new Scanner(System.out);

C. Scanner input = new Scanner(System.err);

D. Scanner input = new Scanner(System.in.read());

Solution

1. Step 1: Identify correct Scanner constructor

   The Scanner constructor requires an InputStream to read input. System.in is the standard input stream (keyboard).

2. Step 2: Check other options

   System.out and System.err are output streams, not input. System.in.read() returns an int, not an InputStream.

3. Final Answer:

   Scanner input = new Scanner(System.in); -> Option A

4. Quick Check:

   Use System.in for input Scanner [OK]

Hint: Use System.in to read keyboard input with Scanner [OK]

Common Mistakes:

• Using System.out or System.err instead of System.in
• Passing System.in.read() which is an int, not InputStream
• Missing 'new' keyword

3. What will be the output of the following code if the user inputs the number 25?

import java.util.Scanner;
public class Test {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int num = sc.nextInt();
    System.out.println(num * 2);
    sc.close();
  }
}

medium

A. Error at runtime

B. 25

C. 50

D. 0

Solution

1. Step 1: Understand input reading

   The program reads an integer from the user using nextInt(). User inputs 25, so num = 25.

2. Step 2: Calculate output

   The program prints num * 2, which is 25 * 2 = 50.

3. Final Answer:

   50 -> Option C

4. Quick Check:

   Input 25 * 2 = 50 [OK]

Hint: Multiply input number by 2 to get output [OK]

Common Mistakes:

• Confusing input with output
• Forgetting to multiply by 2
• Not closing Scanner (though not causing output error)

4. Identify the error in the following code snippet:

import java.util.Scanner;
public class Demo {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter your name:");
    String name = sc.nextInt();
    System.out.println("Hello, " + name);
    sc.close();
  }
}

medium

A. Not closing the Scanner

B. Using nextInt() to read a String

C. Missing import statement for Scanner

D. Incorrect print statement syntax

Solution

1. Step 1: Check input method used

   The code uses nextInt() to read the user's name, which is a String, not an integer.

2. Step 2: Identify correct method

   To read a String, nextLine() or next() should be used instead of nextInt().

3. Final Answer:

   Using nextInt() to read a String -> Option B

4. Quick Check:

   Use nextLine() for String input, not nextInt() [OK]

Hint: Use nextLine() for strings, nextInt() for numbers [OK]

Common Mistakes:

• Using nextInt() for String input
• Ignoring Scanner import
• Not closing Scanner (not error here)
• Misunderstanding print syntax

5. You want to read a full line of text including spaces from the user after reading an integer using Scanner. Which sequence of method calls correctly handles this without skipping input?

Scanner sc = new Scanner(System.in);
int age = sc.nextInt();
// What next?

hard

A. sc.nextLine(); String name = sc.nextLine();

B. String name = sc.nextLine();

C. sc.next(); String name = sc.nextLine();

D. sc.nextInt(); String name = sc.nextLine();

Solution

1. Step 1: Understand Scanner behavior after nextInt()

   After nextInt(), the newline character remains in the input buffer, so a direct nextLine() reads that leftover newline.

2. Step 2: Consume leftover newline before reading full line

   Calling sc.nextLine() once consumes the leftover newline, then the next nextLine() reads the actual full line input.

3. Final Answer:

   sc.nextLine(); String name = sc.nextLine(); -> Option A

4. Quick Check:

   Consume newline with nextLine() before reading full line [OK]

Hint: Call nextLine() twice after nextInt() to read full line [OK]

Common Mistakes:

• Skipping the first nextLine() to consume newline
• Using next() which reads only one word
• Calling nextInt() again unnecessarily