When with ranges and types in Kotlin - Time & Space Complexity

Time Complexity: When with ranges and types

O(1)

Understanding Time Complexity

We want to understand how the time it takes to run a Kotlin when expression changes as the input grows.

Specifically, how checking ranges and types inside when affects performance.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fun checkValue(x: Any) = when (x) {
    in 1..10 -> "Small number"
    in 11..100 -> "Medium number"
    is String -> "A string"
    else -> "Other"
}
    

This code checks if x falls in certain number ranges or is a string, then returns a message.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The when expression checks each condition one by one.
• How many times: It stops as soon as one condition matches, so in the worst case it checks all conditions once.

How Execution Grows With Input

Each check is simple and does not depend on input size. The number of checks is fixed (4 conditions).

Input Size (n)	Approx. Operations
10	4 checks max
100	4 checks max
1000	4 checks max

Pattern observation: The number of operations stays the same no matter how big the input value is.

Final Time Complexity

Time Complexity: O(1)

This means the time to check the value does not grow as the input changes; it stays constant.

Common Mistake

[X] Wrong: "Checking ranges inside when takes longer if the number is bigger."

[OK] Correct: The code only checks fixed ranges one by one, so the size of the number does not affect how many checks happen.

Interview Connect

Understanding how when works with ranges and types helps you explain how your code runs efficiently and predictably, a skill valued in real projects.

Self-Check

What if we added 100 more range checks inside the when? How would the time complexity change?

Practice

1. What does the when expression in Kotlin do when used with ranges like in 1..5?

easy

A. It throws an error if the value is outside the range.

B. It creates a new range from 1 to 5.

C. It assigns the value to the range variable.

D. It checks if the value falls within the specified range.

Solution

1. Step 1: Understand the when with range syntax

   The syntax in 1..5 inside when checks if the value is between 1 and 5 inclusive.

2. Step 2: Interpret the check

   If the value is inside the range, the corresponding branch executes; otherwise, it moves to the next case.

3. Final Answer:

   It checks if the value falls within the specified range. -> Option D

4. Quick Check:

   when with in checks range membership [OK]

Hint: Remember: in means inside the range [OK]

Common Mistakes:

• Thinking in creates a range instead of checking membership
• Confusing in with assignment
• Assuming it throws errors for out-of-range values

2. Which of the following is the correct syntax to check if a variable x is NOT in the range 10 to 20 using when in Kotlin?

easy

A. when(x) { !in 10..20 -> "Outside" }

B. when(x) { not in 10..20 -> "Outside" }

C. when(x) { x !in 10..20 -> "Outside" }

D. when(x) { outside 10..20 -> "Outside" }

Solution

1. Step 1: Recall the correct negation syntax in when

   In Kotlin, !in is used to check if a value is NOT in a range.

2. Step 2: Match the syntax with options

   when(x) { !in 10..20 -> "Outside" } uses !in 10..20 correctly inside when. Other options use invalid keywords or syntax.

3. Final Answer:

   when(x) { !in 10..20 -> "Outside" } -> Option A

4. Quick Check:

   Use !in to check exclusion from range [OK]

Hint: Use !in to say NOT in range [OK]

Common Mistakes:

• Using 'not in' instead of '!in'
• Writing variable name inside when case
• Using undefined keywords like 'outside'

3. What will be the output of this Kotlin code?

val x: Any = 42
val result = when (x) {
  is Int -> "Integer"
  in 1..10 -> "Small number"
  else -> "Unknown"
}
println(result)

medium

A. Small number

B. Unknown

C. Integer

D. Compilation error

Solution

1. Step 1: Check the type of x

   Variable x is of type Any but holds the value 42, which is an Int.

2. Step 2: Evaluate when branches in order

   The first branch checks is Int, which is true, so it returns "Integer". The second branch is not checked because the first matched.

3. Final Answer:

   Integer -> Option C

4. Quick Check:

   when matches first true branch [OK]

Hint: Type checks with is come before range checks [OK]

Common Mistakes:

• Assuming range check runs before type check
• Confusing value 42 as outside 1..10
• Thinking when checks all branches

4. Identify the error in this Kotlin when expression:

val y: Any = "Hello"
val output = when (y) {
  in 1..5 -> "Number"
  is String -> "Text"
  else -> "Other"
}
println(output)

medium

A. Cannot use in with Any type without casting

B. The is String check should come before the range check

C. Missing break statements in when

D. No error, code runs fine

Solution

1. Step 1: Analyze the in 1..5 check on Any

   The variable y is of type Any. The in operator with ranges requires a comparable type like Int. Using in 1..5 on Any causes a type mismatch error.

2. Step 2: Check order of branches and syntax

   The is String check is valid but unreachable if the first branch errors. Also, when in Kotlin does not require break.

3. Final Answer:

   Cannot use in with Any type without casting -> Option A

4. Quick Check:

   Range checks need compatible types [OK]

Hint: Check variable type before using in with ranges [OK]

Common Mistakes:

• Ignoring type mismatch for in operator
• Thinking when needs break
• Assuming order of branches causes error

5. You want to write a Kotlin when expression that returns:
- "Small Int" if the value is an Int between 1 and 10 inclusive,
- "Large Int" if the value is an Int greater than 10,
- "String" if the value is a String,
- "Other" for anything else.

Which of the following when expressions correctly implements this?

hard

A. when (v) {
in 1..10 -> "Small Int"
is Int -> "Large Int"
is String -> "String"
else -> "Other"
}

B. when (v) {
is Int -> if (v in 1..10) "Small Int" else "Large Int"
is String -> "String"
else -> "Other"
}

C. when (v) {
is Int in 1..10 -> "Small Int"
is Int -> "Large Int"
is String -> "String"
else -> "Other"
}

D. when (v) {
is Int && v in 1..10 -> "Small Int"
is Int -> "Large Int"
is String -> "String"
else -> "Other"
}

Solution

1. Step 1: Understand the conditions needed

   We need to check if v is an Int and then check its value range for "Small Int" or "Large Int". Also check if v is a String.

2. Step 2: Analyze each option's correctness

   when (v) { in 1..10 -> "Small Int" is Int -> "Large Int" is String -> "String" else -> "Other" } incorrectly uses in 1..10 without checking type first, causing errors if v is not Int. when (v) { is Int in 1..10 -> "Small Int" is Int -> "Large Int" is String -> "String" else -> "Other" } uses invalid syntax is Int in 1..10. when (v) { is Int && v in 1..10 -> "Small Int" is Int -> "Large Int" is String -> "String" else -> "Other" } uses invalid syntax is Int && v in 1..10 inside when. when (v) { is Int -> if (v in 1..10) "Small Int" else "Large Int" is String -> "String" else -> "Other" } correctly checks type first, then uses an if inside the branch to distinguish ranges.

3. Final Answer:

   Option B correctly implements the logic. -> Option B

4. Quick Check:

   Use nested if inside when for complex conditions [OK]

Hint: Check type first, then range inside branch with if [OK]

Common Mistakes:

• Using invalid syntax combining is and in
• Not checking type before range
• Trying to use logical operators directly in when cases