Partition for splitting in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Kotlin partition example?

Consider the following Kotlin code that uses the partition function to split a list of integers into even and odd numbers. What will be printed?

Kotlin

val numbers = listOf(1, 2, 3, 4, 5, 6)
val (evens, odds) = numbers.partition { it % 2 == 0 }
println("Evens: $evens")
println("Odds: $odds")

A Evens: [1, 3, 5]\nOdds: [2, 4, 6]

B Evens: [2, 4, 6]\nOdds: [1, 3, 5]

C Evens: [1, 2, 3]\nOdds: [4, 5, 6]

D Evens: []\nOdds: [1, 2, 3, 4, 5, 6]

💡 Hint

Remember, partition splits the list into two lists: one where the condition is true, and one where it is false.

The partition function returns a pair of lists. The first list contains elements where the predicate is true (even numbers), and the second list contains elements where the predicate is false (odd numbers).

❓ Predict Output

intermediate

What is the size of the lists after partition?

Given this Kotlin code, how many elements are in the first and second lists after partitioning by the condition it > 3?

Kotlin

val data = listOf(2, 3, 4, 5, 6)
val (greaterThanThree, notGreaterThanThree) = data.partition { it > 3 }
println("greaterThanThree size: ${greaterThanThree.size}")
println("notGreaterThanThree size: ${notGreaterThanThree.size}")

A greaterThanThree size: 3\nnotGreaterThanThree size: 2

B greaterThanThree size: 0\nnotGreaterThanThree size: 5

C greaterThanThree size: 5\nnotGreaterThanThree size: 0

D greaterThanThree size: 2\nnotGreaterThanThree size: 3

💡 Hint

Count how many numbers in the list are greater than 3.

Numbers greater than 3 are 4, 5, and 6 (3 elements). The rest (2 and 3) are not greater than 3 (2 elements).

❓ Predict Output

advanced

What is the output when partitioning a list of strings by length?

Look at this Kotlin code that partitions a list of strings by whether their length is greater than 3. What will be printed?

Kotlin

val words = listOf("cat", "house", "dog", "elephant")
val (longWords, shortWords) = words.partition { it.length > 3 }
println("Long words: $longWords")
println("Short words: $shortWords")

A Long words: []\nShort words: [cat, house, dog, elephant]

B Long words: [cat, dog]\nShort words: [house, elephant]

C Long words: [cat, house, dog]\nShort words: [elephant]

D Long words: [house, elephant]\nShort words: [cat, dog]

💡 Hint

Check the length of each word carefully.

Words with length greater than 3 are "house" (5) and "elephant" (8). The others have length 3 or less.

❓ Predict Output

advanced

What error occurs when using partition incorrectly?

What error will this Kotlin code produce?

Kotlin

val numbers = listOf(1, 2, 3)
val (evens, odds) = numbers.partition { it % 2 == 0 }
println(evens[3])

A NoSuchElementException

B NullPointerException

C IndexOutOfBoundsException

D ClassCastException

💡 Hint

Check the size of the evens list and the index accessed.

The evens list contains [2]. Accessing index 3 is out of bounds, causing IndexOutOfBoundsException.

🧠 Conceptual

expert

Which option correctly describes the behavior of Kotlin's partition function?

Choose the statement that best describes what Kotlin's partition function does when called on a collection.

A It splits the collection into two lists: the first contains elements matching the predicate, the second contains elements not matching it.

B It filters the collection and returns a single list with elements matching the predicate.

C It removes all elements that match the predicate from the original collection.

D It sorts the collection into ascending and descending order lists.

💡 Hint

Think about how partition divides elements based on a condition.

Partition returns a pair of lists: one with elements where the predicate is true, and one where it is false.

Practice

1. What does the partition function do in Kotlin when applied to a list?

easy

A. Converts the list into a map with indices as keys.

B. Sorts the list in ascending order.

C. Removes duplicate elements from the list.

D. Splits the list into two lists based on a condition, returning a Pair of lists.

Solution

1. Step 1: Understand the purpose of partition

   The partition function divides a list into two parts based on a condition.

2. Step 2: Recognize the return type

   It returns a Pair where the first list contains elements matching the condition, and the second list contains the rest.

3. Final Answer:

   Splits the list into two lists based on a condition, returning a Pair of lists. -> Option D

4. Quick Check:

   partition = split by condition [OK]

Hint: Remember: partition always returns two lists in a Pair [OK]

Common Mistakes:

• Thinking partition sorts the list
• Confusing partition with filtering
• Assuming partition removes duplicates

2. Which of the following is the correct syntax to use partition on a list numbers to split even and odd numbers?

easy

A. val (evens, odds) = numbers.partition { it % 2 == 0 }

B. val (evens, odds) = numbers.partition(it % 2 == 0)

C. val evens = numbers.partition { it % 2 == 0 }

D. val evens, odds = numbers.partition { it % 2 == 0 }

Solution

1. Step 1: Check the lambda syntax inside partition

   The condition must be inside curly braces with it representing each element.

2. Step 2: Verify destructuring declaration

   Partition returns a Pair, so use val (first, second) = ... to get both lists.

3. Final Answer:

   val (evens, odds) = numbers.partition { it % 2 == 0 } -> Option A

4. Quick Check:

   Correct lambda and destructuring = val (evens, odds) = numbers.partition { it % 2 == 0 } [OK]

Hint: Use val (a, b) = list.partition { condition } [OK]

Common Mistakes:

• Missing curly braces for lambda
• Not using destructuring for Pair
• Using parentheses instead of braces for lambda

3. What is the output of this Kotlin code?

val words = listOf("apple", "banana", "avocado", "berry")
val (aWords, otherWords) = words.partition { it.startsWith('a') }
println(aWords)
println(otherWords)

medium

A. ["apple", "avocado"]
["banana", "berry"]

B. ["banana", "berry"]
["apple", "avocado"]

C. ["apple"]
["banana", "avocado", "berry"]

D. ["avocado"]
["apple", "banana", "berry"]

Solution

1. Step 1: Apply the condition to each word

   Check which words start with 'a': "apple" and "avocado" match.

2. Step 2: Partition splits into two lists

   First list has words starting with 'a', second list has the rest: "banana", "berry".

3. Final Answer:

   ["apple", "avocado"] ["banana", "berry"] -> Option A

4. Quick Check:

   partition by startsWith('a') = ["apple", "avocado"] ["banana", "berry"] [OK]

Hint: Check condition on each element carefully [OK]

Common Mistakes:

• Mixing order of lists in output
• Forgetting that partition returns Pair of lists
• Confusing startsWith with contains

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

val nums = listOf(1, 2, 3, 4)
val evens, odds = nums.partition { it % 2 == 0 }
println(evens)
println(odds)

medium

A. partition cannot be used on lists of integers.

B. The lambda condition is incorrect; should be it % 2 != 0.

C. Missing parentheses around evens and odds for destructuring declaration.

D. println statements are missing string interpolation.

Solution

1. Step 1: Check destructuring declaration syntax

   Destructuring a Pair requires parentheses: val (evens, odds) = ....

2. Step 2: Confirm lambda and usage

   The lambda is correct for evens; partition works on lists of integers; println is fine.

3. Final Answer:

   Missing parentheses around evens and odds for destructuring declaration. -> Option C

4. Quick Check:

   Destructuring needs parentheses [OK]

Hint: Use val (a, b) = ... for destructuring Pair [OK]

Common Mistakes:

• Omitting parentheses in destructuring
• Changing correct lambda unnecessarily
• Thinking partition only works on strings

5. Given a list of mixed integers including zero and negatives, how can you use partition to separate positive numbers from zero and negatives? Choose the correct code snippet.

hard

A. val (positives, nonPositives) = numbers.partition { it < 0 }

B. val (positives, nonPositives) = numbers.partition { it > 0 }

C. val (positives, nonPositives) = numbers.partition { it >= 0 }

D. val (positives, nonPositives) = numbers.partition { it == 0 }

Solution

1. Step 1: Define the condition for positives

   Positive numbers are greater than zero, so condition is it > 0.

2. Step 2: Use partition with this condition

   Partition splits list into positives and the rest (zero and negatives).

3. Final Answer:

   val (positives, nonPositives) = numbers.partition { it > 0 } -> Option B

4. Quick Check:

   Partition by it > 0 separates positives [OK]

Hint: Use it > 0 to get positives, rest are zero/negatives [OK]

Common Mistakes:

• Using it >= 0 includes zero as positive
• Using it < 0 reverses groups
• Using it == 0 only selects zeros