Number literal formats (underscore, hex, binary) in Kotlin - Time & Space Complexity

Time Complexity: Number literal formats (underscore, hex, binary)

O(n)

Understanding Time Complexity

We want to understand how the time it takes to read and use number literals changes as the numbers get bigger or more complex.

How does the format of a number affect the work the program does?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

val decimal = 1_000_000
val hex = 0xFF_EC_DE_5E
val binary = 0b11010010_01101001_10010100_10010010

fun printNumbers() {
    println(decimal)
    println(hex)
    println(binary)
}
    

This code defines numbers using decimal with underscores, hexadecimal, and binary formats, then prints them.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Reading and converting the number literals once when the program starts.
• How many times: Each number literal is processed only once during initialization.

How Execution Grows With Input

As the number literals get longer (more digits or bits), the time to read and convert them grows a little.

Input Size (digits/bits)	Approx. Operations
10	About 10 steps to read and convert
100	About 100 steps to read and convert
1000	About 1000 steps to read and convert

Pattern observation: The work grows directly with the length of the number literal.

Final Time Complexity

Time Complexity: O(n)

This means the time to process number literals grows in a straight line with the number of digits or bits.

Common Mistake

[X] Wrong: "Using underscores or hex/binary formats makes the program slower in a big way."

[OK] Correct: The program only reads and converts these numbers once, so the difference is very small and grows only with the length of the number.

Interview Connect

Understanding how number formats affect processing helps you explain how programs handle data efficiently, a useful skill in many coding situations.

Self-Check

"What if we used very large number literals repeatedly inside a loop? How would the time complexity change?"

Practice

1. Which of the following is a valid way to write the number one million in Kotlin using underscores for readability?

easy

A. 1__000000

B. 1000_000_

C. 10__00_000

D. 1_000_000

Solution

1. Step 1: Understand underscore usage in Kotlin numbers

   Kotlin allows underscores between digits to improve readability, but they must not be consecutive, leading, or trailing.

2. Step 2: Check each option for correct underscore placement

   1_000_000 uses single underscores separating groups of three digits correctly. A has consecutive underscores (invalid). B has trailing underscore (invalid). C has consecutive underscores (invalid).

3. Final Answer:

   1_000_000 -> Option D

4. Quick Check:

   Underscores separate digits properly = 1_000_000 [OK]

Hint: Use single underscores between digit groups only [OK]

Common Mistakes:

• Using double underscores
• Placing underscores at start or end
• Trailing underscores

2. Which of the following Kotlin number literals is NOT valid syntax?

easy

A. 123_456_

B. 0b1010_1101

C. 0x1A3F

D. 987_654

Solution

1. Step 1: Review Kotlin rules for underscores in number literals

   Underscores can be used between digits but not at the end or start of a number.

2. Step 2: Check each option for trailing underscores

   123_456_ ends with an underscore, which is invalid syntax. Options A, B, and D are valid hexadecimal, binary, and decimal literals respectively.

3. Final Answer:

   123_456_ -> Option A

4. Quick Check:

   Trailing underscore invalid = 123_456_ [OK]

Hint: Never put underscore at the end of a number [OK]

Common Mistakes:

• Trailing underscores in numbers
• Confusing hex prefix 0x with decimal
• Misplacing underscores in binary literals

3. What is the output of this Kotlin code?

val x = 0xF_A
val y = 0b1010_1010
println(x + y)

medium

A. 270

B. 250

C. 420

D. 280

Solution

1. Step 1: Convert hexadecimal and binary literals to decimal

   0xF_A means hex FA: F=15, A=10, so 15*16 + 10 = 250. 0b1010_1010 is binary 10101010 which equals 128+32+8+2=170.

2. Step 2: Add the decimal values

   250 + 170 = 420.

3. Final Answer:

   420 -> Option C

4. Quick Check:

   (0xFA + 0b10101010) = 420 [OK]

Hint: Convert hex and binary to decimal before adding [OK]

Common Mistakes:

• Misreading hex digits
• Incorrect binary to decimal conversion
• Ignoring underscores in literals

4. Identify the error in this Kotlin code snippet:

val num = 0b1021_0010

medium

A. Underscores are not allowed in binary literals

B. Binary literals cannot contain digits other than 0 or 1

C. Binary literals must start with 0x

D. The number is too large for binary format

Solution

1. Step 1: Understand valid digits in binary literals

   Binary literals can only contain digits 0 and 1, optionally separated by underscores.

2. Step 2: Check the given literal for invalid digits

   The literal contains '2', which is invalid in binary. Underscores are allowed, and binary literals start with 0b, not 0x.

3. Final Answer:

   Binary literals cannot contain digits other than 0 or 1 -> Option B

4. Quick Check:

   Binary digits must be 0 or 1 only [OK]

Hint: Binary digits only 0 or 1, no other digits allowed [OK]

Common Mistakes:

• Using digits other than 0 or 1 in binary
• Confusing hex prefix 0x with binary 0b
• Thinking underscores are disallowed in binary

5. You want to represent the decimal number 255 in Kotlin using a binary literal with underscores for readability. Which of the following is the best way to write it?

hard

A. 0b1111_1111

B. 0b11111111

C. 0b1111_1110

D. 0b1111_11112

Solution

1. Step 1: Convert decimal 255 to binary

   255 in binary is 11111111 (eight ones).

2. Step 2: Add underscores for readability

   Grouping bits in fours is common, so 1111_1111 is clearer and valid.

3. Step 3: Check options for correctness

   0b1111_1111 correctly uses underscores and matches 255. 0b11111111 is correct but less readable. 0b1111_1110 is 254, and 0b1111_11112 has invalid digit '2'.

4. Final Answer:

   0b1111_1111 -> Option A

5. Quick Check:

   Binary 1111_1111 = 255 decimal [OK]

Hint: Group binary digits in fours with underscores [OK]

Common Mistakes:

• Using wrong binary digits
• Adding invalid digits like 2
• Not grouping bits for readability