Sequence creation methods in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Kotlin sequence code?

Consider the following Kotlin code that creates and processes a sequence. What will be printed?

Kotlin

val seq = sequenceOf(1, 2, 3, 4, 5).filter { it % 2 == 0 }.map { it * 3 }
println(seq.toList())

A [2, 4, 6, 8, 10]

B [6, 12]

C [3, 6, 9, 12, 15]

D [1, 2, 3, 4, 5]

💡 Hint

Remember that filter keeps only even numbers, then map multiplies them by 3.

The sequence filters even numbers (2,4) and then multiplies each by 3, resulting in [6,12].

🧠 Conceptual

intermediate

Which Kotlin sequence creation method produces an infinite sequence?

Among the following Kotlin sequence creation methods, which one can produce an infinite sequence?

A generateSequence { (1..10).random() }

B sequenceOf(1, 2, 3)

C listOf(1, 2, 3).asSequence()

D emptySequence<Int>()

💡 Hint

Think about which method can keep producing values without a defined end.

generateSequence with a lambda that always returns a value can produce an infinite sequence.

🔧 Debug

advanced

What error does this Kotlin sequence code raise?

Examine this Kotlin code snippet. What error will it cause when run?

Kotlin

val seq = generateSequence(1) { if (it < 5) it + 1 else null }
println(seq.toList())

A No error, prints [1, 2, 3, 4, 5]

B Runtime error: NullPointerException

C Compilation error: Missing else branch in lambda

D Infinite loop causing OutOfMemoryError

💡 Hint

Check the lambda's return type and what happens when condition is false.

The lambda must return null to end the sequence, and here it returns null when condition is false, so no error occurs.

📝 Syntax

advanced

Which option correctly creates a Kotlin sequence from a range and filters odd numbers?

Select the Kotlin code snippet that correctly creates a sequence from 1 to 10 and filters only odd numbers.

A val seq = generateSequence(1..10) { it % 2 == 1 }

B val seq = sequenceOf(1..10).filter { it % 2 == 1 }

C val seq = (1..10).asSequence().filter { it % 2 == 1 }

D val seq = (1..10).filter { it % 2 == 1 }.sequence()

💡 Hint

Remember how to convert a range to a sequence and apply filter.

Option C correctly converts the range to a sequence and filters odd numbers. Others have syntax or logic errors.

🚀 Application

expert

How many items are in the resulting sequence?

Given this Kotlin code, how many items will the resulting sequence contain?

Kotlin

val seq = generateSequence(1) { if (it < 10) it + 2 else null }
val result = seq.toList()

A 9

B 6

C 10

D 5

💡 Hint

Count the numbers starting at 1, adding 2 each time, stopping before exceeding 10.

Sequence generates: 1, 3, 5, 7, 9 and stops because next would be 11 which is > 10. So 5 items total.