Java streams vs Kotlin sequences - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a Kotlin sequence from a list.

Kotlin

val numbers = listOf(1, 2, 3, 4, 5)
val seq = numbers.[1]()

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

AtoSequence

Bstream

CasSequence

DsequenceOf

Attempts:

3 left

💡 Hint

Common Mistakes

Using Java's stream() method which is not available on Kotlin lists.

Trying to use toSequence() which does not exist.

Use asSequence() to convert a list to a Kotlin sequence.

2fill in blank

medium

Complete the code to filter even numbers using a Kotlin sequence.

Kotlin

val numbers = listOf(1, 2, 3, 4, 5)
val evens = numbers.asSequence().[1] { it % 2 == 0 }.toList()

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

AflatMap

Bmap

Creduce

Dfilter

Attempts:

3 left

💡 Hint

Common Mistakes

Using map which transforms elements but does not filter.

Using reduce which combines elements into one value.

The filter function selects elements matching the condition.

3fill in blank

hard

Fix the error in the code to convert a Kotlin sequence back to a list.

Kotlin

val seq = listOf(1, 2, 3).asSequence()
val list = seq.[1]()

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

AasList

BtoList

CtoSequence

DtoArray

Attempts:

3 left

💡 Hint

Common Mistakes

Using toSequence() which does not convert back to list.

Using asList() which is not a Kotlin function.

Use toList() to convert a sequence back to a list.

4fill in blank

hard

Fill both blanks to create a Java stream from a list and collect it back to a list.

Kotlin

val list = listOf(1, 2, 3, 4)
val stream = list.[1]()
val result = stream.collect(Collectors.[2]())

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

Astream

BtoList

DtoSet

Attempts:

3 left

💡 Hint

Common Mistakes

Using Kotlin sequence methods instead of Java stream methods.

Using toSet collector instead of toList.

Use stream() to get a Java stream and Collectors.toList() to collect it back to a list.

5fill in blank

hard

Fill all three blanks to create a Kotlin sequence, filter odd numbers, and convert to a list.

Kotlin

val numbers = listOf(1, 2, 3, 4, 5)
val result = numbers.[1]().[2] { it % 2 != 0 }.[3]()

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

AasSequence

Bfilter

CtoList

Dmap

Attempts:

3 left

💡 Hint

Common Mistakes

Using map instead of filter for selecting odd numbers.

Forgetting to convert the sequence back to a list.

First convert to a sequence with asSequence(), then filter odd numbers with filter, and finally convert back to a list with toList().

Practice

1. Which statement best describes the main similarity between Java streams and Kotlin sequences?

easy

A. Both provide lazy evaluation to process collections efficiently.

B. Both are eager and process all elements immediately.

C. Both are only available in Kotlin.

D. Both require manual iteration over collections.

Solution

1. Step 1: Understand lazy evaluation concept

   Lazy evaluation means processing elements only when needed, not all at once.

2. Step 2: Compare Java streams and Kotlin sequences

   Both Java streams and Kotlin sequences use lazy evaluation to improve performance when processing collections.

3. Final Answer:

   Both provide lazy evaluation to process collections efficiently. -> Option A

4. Quick Check:

   Lazy evaluation = Both streams and sequences [OK]

Hint: Remember: streams and sequences delay processing until needed [OK]

Common Mistakes:

• Thinking both are eager by default
• Assuming Kotlin sequences are only in Java
• Believing manual iteration is required

2. Which of the following is the correct way to create a Kotlin sequence from a list?

easy

A. val seq = list.stream()

B. val seq = list.toStream()

C. val seq = Sequence(list)

D. val seq = list.asSequence()

Solution

1. Step 1: Recall Kotlin sequence creation syntax

   In Kotlin, you convert a collection to a sequence using the asSequence() extension function.

2. Step 2: Check each option

   val seq = list.stream() uses Java stream syntax, val seq = Sequence(list) is invalid constructor usage, val seq = list.toStream() is not a Kotlin function.

3. Final Answer:

   val seq = list.asSequence() -> Option D

4. Quick Check:

   Use asSequence() to get Kotlin sequence [OK]

Hint: Use asSequence() to convert collections in Kotlin [OK]

Common Mistakes:

• Using Java stream() method in Kotlin
• Trying to instantiate Sequence directly
• Using non-existent toStream() method

3. What is the output of the following Kotlin code?

val list = listOf(1, 2, 3, 4)
val result = list.asSequence()
    .filter { it % 2 == 0 }
    .map { it * 2 }
    .toList()
println(result)

medium

A. [4, 8]

B. [1, 2, 3, 4]

C. [1, 3]

D. [2, 4, 6, 8]

Solution

1. Step 1: Apply filter to keep even numbers

   Filter keeps elements where it % 2 == 0, so from [1,2,3,4] we get [2,4].

2. Step 2: Map each filtered element by multiplying by 2

   Mapping [2,4] with it * 2 gives [4,8].

3. Final Answer:

   [4, 8] -> Option A

4. Quick Check:

   Filter evens then double = [4, 8] [OK]

Hint: Filter first, then map for correct sequence output [OK]

Common Mistakes:

• Mapping before filtering
• Confusing filter condition
• Expecting original list output

4. Identify the error in this Kotlin code using sequences:

val numbers = listOf(1, 2, 3)
val seq = numbers.asSequence()
val result = seq.filter { it > 1 }
println(result)

medium

A. Filter condition syntax is invalid.

B. Incorrect use of asSequence() on a list.

C. Missing terminal operation to trigger sequence evaluation.

D. Cannot print a sequence directly.

Solution

1. Step 1: Understand sequence evaluation

   Kotlin sequences are lazy and require a terminal operation like toList() to execute.

2. Step 2: Check code for terminal operation

   The code filters but does not collect or convert the sequence, so result is still a sequence, not a list.

3. Final Answer:

   Missing terminal operation to trigger sequence evaluation. -> Option C

4. Quick Check:

   Sequences need terminal operation to run [OK]

Hint: Always end sequences with terminal operation like toList() [OK]

Common Mistakes:

• Assuming filter runs immediately
• Trying to print sequence without conversion
• Confusing asSequence() usage

5. You want to process a large list of numbers in Kotlin, doubling only the odd numbers, but you want to avoid processing the entire list eagerly. Which approach is best?

1) Use list.map { if (it % 2 != 0) it * 2 else it }
2) Use list.asSequence().filter { it % 2 != 0 }.map { it * 2 }.toList()
3) Use list.stream().filter { it % 2 != 0 }.map { it * 2 }.collect(Collectors.toList())
4) Use list.asSequence().map { if (it % 2 != 0) it * 2 else it }.toList()

hard

A. Option 1 because map is simple and eager.

B. Option 2 because sequences filter and map lazily, processing only needed elements.

C. Option 3 because Java streams are faster in Kotlin.

D. Option 4 because it maps lazily but does not filter.

Solution

1. Step 1: Understand lazy processing need

   To avoid eager processing of the entire list, lazy evaluation is needed.

2. Step 2: Analyze each option's behavior

   Option 1 is eager map, processes all elements. Option 2 uses Kotlin sequences with filter and map lazily, then collects. Option 3 uses Java streams but in Kotlin context, less idiomatic. Option 4 maps lazily but does not filter, so processes all elements (conditionally doubling only odds).

3. Step 3: Choose best approach

   Option 2 best fits lazy processing of only odd numbers, avoiding unnecessary work.

4. Final Answer:

   Option 2 because sequences filter and map lazily, processing only needed elements. -> Option B

5. Quick Check:

   Lazy filter + map with sequences = Option 2 [OK]

Hint: Use sequences with filter then map for lazy selective processing [OK]

Common Mistakes:

• Using eager map on entire list
• Ignoring lazy evaluation benefits
• Confusing Java streams with Kotlin sequences