Range operator (..) and in operator in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of range and in operator with characters

What is the output of this Kotlin code snippet?

Kotlin

val result = 'c' in 'a'..'f'
println(result)

A false

B Compilation error

C Runtime exception

D true

💡 Hint

Check if the character 'c' lies between 'a' and 'f' inclusive.

The range 'a'..'f' includes all characters from 'a' to 'f'. Since 'c' is within this range, the expression returns true.

❓ Predict Output

intermediate

Range operator with integers and in operator

What will be printed by this Kotlin code?

Kotlin

val x = 10
val isInRange = x in 1..10
println(isInRange)

A true

B false

C Compilation error

D Runtime exception

💡 Hint

Remember the range 1..10 includes both 1 and 10.

The number 10 is included in the range 1..10, so x in 1..10 evaluates to true.

❓ Predict Output

advanced

Output of reversed range with in operator

What is the output of this Kotlin code?

Kotlin

val num = 5
val range = 10 downTo 1
println(num in range)

A false

B true

C Compilation error

D Runtime exception

💡 Hint

Check if 5 is within the range from 10 down to 1.

The downTo operator creates a decreasing range from 10 to 1. Since 5 is between 10 and 1, num in range is true.

❓ Predict Output

advanced

Checking if a value is in a step range

What will this Kotlin code print?

Kotlin

val range = 1..10 step 3
println(7 in range)

A Compilation error

B true

C false

D Runtime exception

💡 Hint

The range includes numbers 1, 4, 7, 10 or not?

The range 1..10 step 3 includes 1, 4, 7, and 10. However, the in operator checks for containment based on the step sequence. Since 7 is included in the stepped range, the output should be true. But in Kotlin, the in operator with stepped ranges checks for exact matches in the progression. Actually, 7 is included, so the output is true.

❓ Predict Output

expert

Output of complex range and in operator with floating point

What is the output of this Kotlin code snippet?

Kotlin

val x = 3.5
val range = 1.0..5.0
println(x in range)

A Compilation error

B false

C Runtime exception

D true

💡 Hint

Check if the range operator (..) works with Double type in Kotlin.

The range operator (..) in Kotlin does not support floating point types like Double. This code will not compile because 1.0..5.0 is invalid.

Practice

1. What does the Kotlin expression 1..5 represent?

easy

A. An error because ranges need parentheses

B. A list containing numbers 1 and 5 only

C. A function that returns 1 or 5

D. A range of numbers from 1 to 5 inclusive

Solution

1. Step 1: Understand the range operator

   The .. operator creates a range including both start and end values.

2. Step 2: Interpret 1..5

   This means all numbers from 1 up to 5, including 1 and 5.

3. Final Answer:

   A range of numbers from 1 to 5 inclusive -> Option D

4. Quick Check:

   1..5 = numbers 1 to 5 [OK]

Hint: Remember: .. includes both ends in Kotlin ranges [OK]

Common Mistakes:

• Thinking it creates a list instead of a range
• Assuming it excludes the last number
• Confusing it with a function call

2. Which of the following is the correct syntax to check if variable x is between 10 and 20 inclusive in Kotlin?

easy

A. if (x between 10 and 20) { }

B. if (x in 10..20) { }

C. if (10 <= x <= 20) { }

D. if (x inside 10..20) { }

Solution

1. Step 1: Recall Kotlin's in operator syntax

   Kotlin uses in to check if a value is inside a range.

2. Step 2: Match syntax with options

   Only if (x in 10..20) correctly uses in and the range operator ...

3. Final Answer:

   if (x in 10..20) { } -> Option B

4. Quick Check:

   Use in with .. for range checks [OK]

Hint: Use 'in' with '..' for range checks in Kotlin [OK]

Common Mistakes:

• Using 'between' or 'inside' which are not Kotlin keywords
• Trying chained comparisons like 10 <= x <= 20
• Missing the range operator '..'

3. What is the output of this Kotlin code?

val x = 7
if (x in 1..5) println("Inside") else println("Outside")

medium

A. Outside

B. Compilation error

C. Inside

D. No output

Solution

1. Step 1: Understand the range check

   The range 1..5 includes numbers 1, 2, 3, 4, 5.

2. Step 2: Check if x=7 is in the range

   7 is not between 1 and 5, so the condition is false.

3. Final Answer:

   Outside -> Option A

4. Quick Check:

   7 not in 1..5 means print "Outside" [OK]

Hint: Check if value lies within range boundaries [OK]

Common Mistakes:

• Assuming 7 is inside 1..5
• Confusing inclusive ranges with exclusive
• Expecting a syntax error

4. Identify the error in this Kotlin code snippet:

for (i in 1..5)
    if (i in 3..1) println(i)

medium

A. The for loop syntax is incorrect

B. The range 3..1 is invalid because start is greater than end

C. No error, code runs fine

D. The if condition should use 'not in' instead of 'in'

Solution

1. Step 1: Analyze the range 3..1

   In Kotlin, a range with start greater than end (3..1) creates an empty range, which is valid.

2. Step 2: Verify code execution

   The for loop runs for i=1 to 5, but i in 3..1 is always false for any i, so nothing prints. No syntax or runtime errors occur.

3. Final Answer:

   No error, code runs fine -> Option C

4. Quick Check:

   Empty ranges (start > end) are valid in Kotlin [OK]

Hint: Ranges with start > end are empty but valid [OK]

Common Mistakes:

• Thinking reversed ranges like 3..1 are invalid
• Assuming the for loop syntax is wrong
• Expecting a syntax error or crash with empty range

5. You want to print all even numbers between 2 and 10 inclusive using a Kotlin for loop and range. Which code snippet correctly does this?

hard

A. for (i in 2..10 step 2) println(i)

B. for (i in 2..10) if (i % 2 != 0) println(i)

C. for (i in 2 until 10 step 2) println(i)

D. for (i in 2..10 step 3) println(i)

Solution

1. Step 1: Understand the range and step usage

   2..10 step 2 creates numbers 2, 4, 6, 8, 10.

2. Step 2: Check each option

   for (i in 2..10 step 2) println(i) prints 2,4,6,8,10 directly. for (i in 2..10) if (i % 2 != 0) println(i) checks i % 2 != 0 and prints odds 3,5,7,9. for (i in 2 until 10 step 2) println(i) uses until excluding 10 (2,4,6,8). for (i in 2..10 step 3) println(i) steps by 3 (2,5,8).

3. Final Answer:

   for (i in 2..10 step 2) println(i) -> Option A

4. Quick Check:

   Use step 2 with inclusive range for even numbers [OK]

Hint: Use 'step 2' with '..' to get even numbers directly [OK]

Common Mistakes:

• Using wrong modulo condition like % 2 != 0
• Using 'until' which excludes the end value
• Using wrong step size like 3