Sequence creation methods in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a 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'

AasSequence

BtoSequence

CgenerateSequence

DsequenceOf

Attempts:

3 left

💡 Hint

Common Mistakes

Using sequenceOf() which creates a sequence from given elements, not from a list.

Trying to call toSequence() which does not exist.

The asSequence() function converts a collection into a sequence.

2fill in blank

medium

Complete the code to create a sequence of numbers starting from 1.

Kotlin

val seq = [1](1) { it + 1 }

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

AsequenceOf

BlistOf

CasSequence

DgenerateSequence

Attempts:

3 left

💡 Hint

Common Mistakes

Using sequenceOf() which creates a sequence from fixed elements.

Using asSequence() which converts collections but does not generate sequences.

The generateSequence() function creates a sequence by generating values starting from the initial value.

3fill in blank

hard

Fix the error in the code to create a sequence from given elements.

Kotlin

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

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

AsequenceOf

BgenerateSequence

CasSequence

DlistOf

Attempts:

3 left

💡 Hint

Common Mistakes

Using asSequence() which converts collections but does not accept varargs.

Using listOf() which creates a list, not a sequence.

The sequenceOf() function creates a sequence from the given elements.

4fill in blank

hard

Fill both blanks to create a sequence from a list and filter even numbers.

Kotlin

val numbers = listOf(1, 2, 3, 4, 5)
val evenSeq = numbers.[1]().filter { it [2] 2 == 0 }

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

AasSequence

BtoSequence

C%

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using toSequence() which does not exist.

Using '==' instead of '%' for checking even numbers.

Use asSequence() to convert the list, then filter numbers where the remainder when divided by 2 is zero.

5fill in blank

hard

Fill all three blanks to create a sequence of squares of numbers from 1 to 5, filtering those greater than 10.

Kotlin

val seq = (1..5).[1]().map { it [2] it }.filter { it [3] 10 }

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

AasSequence

B*

C>

DsequenceOf

Attempts:

3 left

💡 Hint

Common Mistakes

Using sequenceOf() instead of asSequence() on a range.

Using '+' instead of '*' for squaring.

Using '<' instead of '>' for filtering.

Convert the range to a sequence, map each number to its square using '*', then filter numbers greater than 10.

Practice

1. Which of the following is the correct way to directly create a sequence from a fixed set of values in Kotlin?

easy

A. val seq = sequenceOf(1, 2, 3)

B. val seq = listOf(1, 2, 3).asSequence()

C. val seq = arrayOf(1, 2, 3).toSequence()

D. val seq = sequence(1, 2, 3)

Solution

1. Step 1: Understand sequence creation from fixed values

   The function sequenceOf() creates a sequence directly from given values.

2. Step 2: Check each option's validity

   val seq = sequenceOf(1, 2, 3) uses sequenceOf(1, 2, 3), which is correct. val seq = listOf(1, 2, 3).asSequence() creates a list then converts to sequence, which is valid but not direct creation. val seq = arrayOf(1, 2, 3).toSequence() uses a non-existent toSequence() method on array, which is invalid. val seq = sequence(1, 2, 3) uses sequence() which is not a valid function for this purpose.

3. Final Answer:

   val seq = sequenceOf(1, 2, 3) -> Option A

4. Quick Check:

   Direct sequence creation = sequenceOf(...) [OK]

Hint: Use sequenceOf() to create sequences from fixed values [OK]

Common Mistakes:

• Using non-existent toSequence() on arrays
• Confusing sequenceOf() with sequence()
• Trying to create sequence directly from array without conversion

2. Which of the following is the correct syntax to create a sequence from a range in Kotlin?

easy

A. val seq = (1..5).toSequence()

B. val seq = sequence(1..5)

C. val seq = (1..5).asSequence()

D. val seq = sequenceOf(1..5)

Solution

1. Step 1: Identify how to convert a range to a sequence

   Ranges in Kotlin can be converted to sequences using the asSequence() extension function.

2. Step 2: Evaluate each option's syntax

   val seq = (1..5).asSequence() uses (1..5).asSequence(), which is correct. val seq = (1..5).toSequence() uses toSequence() which does not exist for ranges. val seq = sequence(1..5) uses sequence(1..5) which is invalid syntax. val seq = sequenceOf(1..5) creates a sequence of a single range object, not elements.

3. Final Answer:

   val seq = (1..5).asSequence() -> Option C

4. Quick Check:

   Range to sequence = asSequence() [OK]

Hint: Use asSequence() to convert ranges to sequences [OK]

Common Mistakes:

• Using toSequence() which does not exist
• Passing range directly to sequence() function
• Using sequenceOf() with a range as single element

3. What will be the output of the following Kotlin code?

val seq = generateSequence(1) { it + 2 }.take(4).toList()
println(seq)

medium

A. [1, 3, 5, 7]

B. [1, 2, 3, 4]

C. [2, 4, 6, 8]

D. [1, 3, 5]

Solution

1. Step 1: Understand generateSequence with lambda

   The sequence starts at 1, then each next element is previous + 2, so elements are 1, 3, 5, 7, ...

2. Step 2: Apply take(4) and convert to list

   Taking first 4 elements gives [1, 3, 5, 7]. Printing this list outputs exactly that.

3. Final Answer:

   [1, 3, 5, 7] -> Option A

4. Quick Check:

   Sequence increments by 2 starting at 1 = [1, 3, 5, 7] [OK]

Hint: generateSequence(start) { it + step } creates arithmetic sequence [OK]

Common Mistakes:

• Confusing take(4) with takeWhile
• Assuming sequence increments by 1 instead of 2
• Forgetting to convert sequence to list before printing

4. Identify the error in the following Kotlin code that tries to create a sequence of even numbers starting from 2:

val seq = generateSequence(2) { it + 2 }
val firstThree = seq.take(3).toList()
println(firstThree)

medium

A. generateSequence requires a nullable lambda return type

B. take() does not exist for sequences

C. toList() cannot be called on a sequence

D. No error; code works correctly

Solution

1. Step 1: Check generateSequence usage

   The code uses generateSequence with a start value and lambda to add 2 each time, which is valid.

2. Step 2: Verify take() and toList() usage

   Sequences support take() to limit elements and toList() to convert to list. Both are valid here.

3. Final Answer:

   No error; code works correctly -> Option D

4. Quick Check:

   generateSequence + take + toList = valid sequence creation [OK]

Hint: generateSequence with non-null lambda works fine [OK]

Common Mistakes:

• Thinking lambda must return nullable type
• Believing toList() is invalid on sequences
• Assuming take() is not available for sequences

5. You want to create a sequence of Fibonacci numbers up to 100 using generateSequence. Which of the following code snippets correctly implements this?

hard

A. val fibSeq = generateSequence(0) { it + it - 1 }
val result = fibSeq.takeWhile { it <= 100 }.toList()

B. val fibSeq = generateSequence(Pair(0, 1)) { (a, b) -> if (b <= 100) Pair(b, a + b) else null }
val result = fibSeq.map { it.first }.toList()

C. val fibSeq = sequenceOf(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144)
val result = fibSeq.filter { it <= 100 }.toList()

D. val fibSeq = generateSequence(0 to 1) { (a, b) -> Pair(b, a + b) }
val result = fibSeq.map { it.second }.takeWhile { it <= 100 }.toList()

Solution

1. Step 1: Understand Fibonacci sequence generation with generateSequence

   We use a Pair to hold two consecutive Fibonacci numbers. The lambda returns next pair if next number is ≤ 100, else null to stop.

2. Step 2: Analyze each option

   val fibSeq = generateSequence(Pair(0, 1)) { (a, b) -> if (b <= 100) Pair(b, a + b) else null }
   val result = fibSeq.map { it.first }.toList()

   correctly uses Pair and stops at 100, then maps to first element for Fibonacci numbers.

   val fibSeq = generateSequence(0) { it + it - 1 }
   val result = fibSeq.takeWhile { it <= 100 }.toList()

   uses incorrect formula.

   val fibSeq = sequenceOf(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144)
   val result = fibSeq.filter { it <= 100 }.toList()

   hardcodes values but includes 144 which is > 100.

   val fibSeq = generateSequence(0 to 1) { (a, b) -> Pair(b, a + b) }
   val result = fibSeq.map { it.second }.takeWhile { it <= 100 }.toList()

   maps to second element and takes while ≤ 100, but first element 0 is missed and sequence never stops (infinite).

3. Final Answer:

   generateSequence(Pair(0, 1)) { (a, b) -> if (b <= 100) Pair(b, a + b) else null }.map { it.first }.toList() -> Option B

4. Quick Check:

   generateSequence with Pair and null stop = correct Fibonacci [OK]

Hint: Use Pair and null to stop generateSequence for Fibonacci [OK]

Common Mistakes:

• Not stopping sequence with null when limit exceeded
• Incorrect Fibonacci formula in lambda
• Including numbers beyond limit in hardcoded sequences