Range operator (..) and in operator in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a range from 1 to 5.

Kotlin

val numbers = 1 [1] 5

Drag options to blanks, or click blank then click option'

Auntil

B...

C..

Dto

Attempts:

3 left

💡 Hint

Common Mistakes

Using '...' which is not a valid range operator in Kotlin.

Using 'until' which excludes the end value.

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

2fill in blank

medium

Complete the code to check if 3 is inside the range from 1 to 5.

Kotlin

val isInRange = 3 [1] 1..5

Drag options to blanks, or click blank then click option'

A==

Bin

Ccontains

Dnot in

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which checks equality, not membership.

Using 'not in' which checks the opposite.

The in operator checks if a value is inside a range.

3fill in blank

hard

Fix the error in the code to check if 10 is not in the range from 1 to 5.

Kotlin

val check = 10 [1] 1..5

Drag options to blanks, or click blank then click option'

A!=

Bin

Ccontains

Dnot in

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' which checks if the value is inside.

Using '!=' which checks inequality, not range membership.

The not in operator checks if a value is outside the range.

4fill in blank

hard

Fill both blanks to create a range from 5 to 10 and check if 7 is inside it.

Kotlin

val range = [1] [2] 10
val result = 7 in range

Drag options to blanks, or click blank then click option'

A5

B..

Cuntil

Dto

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'until' which excludes the end value.

Using 'to' which creates a Pair, not a range.

Use 5 as the start and .. to create an inclusive range to 10.

5fill in blank

hard

Fill all three blanks to create a range from 'a' to 'z' and check if 'm' is inside it.

Kotlin

val letters = [1] [2] [3]
val containsM = 'm' in letters

Drag options to blanks, or click blank then click option'

A'a'

B..

C'z'

Duntil

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'until' which excludes the end character.

Using double quotes instead of single quotes for characters.

Use 'a' as start, .. as range operator, and 'z' as end.

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