When with ranges and types in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of when with Int range check

What is the output of this Kotlin code snippet?

Kotlin

fun main() {
    val x = 15
    val result = when (x) {
        in 1..10 -> "Low"
        in 11..20 -> "Medium"
        else -> "High"
    }
    println(result)
}

A Low

B Medium

C High

D Compilation error

💡 Hint

Check which range the value 15 belongs to.

The value 15 falls within the range 11..20, so the when expression returns "Medium".

❓ Predict Output

intermediate

When with type checking

What will be printed by this Kotlin code?

Kotlin

fun main() {
    val obj: Any = "Hello"
    val output = when (obj) {
        is Int -> "Integer"
        is String -> "String"
        else -> "Unknown"
    }
    println(output)
}

A String

B Integer

C Unknown

D Runtime exception

💡 Hint

Check the type of obj at runtime.

The variable obj holds a String, so the when expression matches is String and returns "String".

❓ Predict Output

advanced

When with mixed ranges and types

What is the output of this Kotlin program?

Kotlin

fun main() {
    val input: Any = 25
    val result = when (input) {
        in 1..10 -> "Small number"
        is Int -> "Integer number"
        in 20..30 -> "Medium number"
        else -> "Other"
    }
    println(result)
}

A Integer number

B Medium number

C Small number

D Other

💡 Hint

Remember that when checks branches in order and stops at the first match.

The input 25 is not in 1..10, but is an Int, so the second branch matches and returns "Integer number". The third branch is never reached.

❓ Predict Output

advanced

When with smart casting and type mismatch

What will this Kotlin code print?

Kotlin

fun main() {
    val data: Any = 5.5
    val output = when (data) {
        is Int -> "Int: ${data + 1}"
        is Double -> "Double: ${data + 1}"
        else -> "Unknown type"
    }
    println(output)
}

A Int: 6

B Compilation error

C Double: 6.5

D Unknown type

💡 Hint

Check the actual type of data and how smart casting works.

data is a Double 5.5, so the when branch is Double matches and adds 1 to print "Double: 6.5".

❓ Predict Output

expert

When with complex type and range conditions

What is the output of this Kotlin code?

Kotlin

fun main() {
    val value: Any = 42
    val result = when (value) {
        is Int -> when (value) {
            in 1..10 -> "Small Int"
            in 11..50 -> "Medium Int"
            else -> "Large Int"
        }
        is String -> "String value"
        else -> "Unknown"
    }
    println(result)
}

A Large Int

B Small Int

C Unknown

D Medium Int

💡 Hint

Look carefully at the nested when and the ranges for the Int value.

value is an Int 42, which falls in 11..50, so the nested when returns "Medium Int".

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