GroupBy for categorization in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to group the list of words by their first letter.

Kotlin

val words = listOf("apple", "banana", "apricot", "blueberry")
val grouped = words.groupBy { it.[1] }
println(grouped)

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

Afirst()

Blength

Clast()

Duppercase()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' instead of 'first()' groups by word length, not first letter.

Using 'last()' groups by last letter, not first.

The groupBy function groups elements by the key returned by the lambda. Using it.first() groups words by their first letter.

2fill in blank

medium

Complete the code to group numbers by their remainder when divided by 3.

Kotlin

val numbers = listOf(1, 2, 3, 4, 5, 6)
val grouped = numbers.groupBy { it [1] 3 }
println(grouped)

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

A%

B*

C/

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '*' or '+' changes the number, not the remainder.

Using '/' gives the quotient, not remainder.

The '%' operator gives the remainder. Grouping by 'it % 3' groups numbers by their remainder when divided by 3.

3fill in blank

hard

Fix the error in the code to group words by their length.

Kotlin

val words = listOf("cat", "dog", "bird")
val grouped = words.groupBy { it.[1] }
println(grouped)

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

Asize

Bcount

Clength

Dlength()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' causes errors; that's for collections like lists, strings use 'length'.

Using 'length()' or 'count'; 'length' is a property (no ()), 'count()' is a function but not standard for length.

In Kotlin, the property 'length' gives the length of a string. 'size' is for collections, 'length()' is incorrect (property, no ()), 'count' is a function.

4fill in blank

hard

Fill both blanks to group words by their last character and filter groups with more than one word.

Kotlin

val words = listOf("apple", "angle", "banana", "bottle", "cat")
val grouped = words.groupBy { it.[1] }.filter { it.value.size [2] 1 }
println(grouped)

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

Alast()

B>

C==

Dfirst()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'first()' groups by first letter, not last.

Using '==' 1 filters groups with exactly one word, not more.

Using 'last()' groups words by their last letter. Filtering with '>' 1 keeps groups with more than one word.

5fill in blank

hard

Fill all three blanks to create a map of words grouped by the lowercase version of their first letter, including only groups with more than two words.

Kotlin

val words = listOf("apple", "apricot", "banana", "blueberry", "avocado", "blackberry")
val grouped = words.groupBy { it.[1].toString().[2]() }.filter { it.value.size [3] 2 }
println(grouped)

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

Afirst()

Bsize

C>

Dlowercase

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' as a function instead of property.

Using '>' incorrectly in filter condition.

Using 'toString()' unnecessarily or missing it.

Use 'first()' to get the first letter, then 'lowercase()' to convert it to lowercase. Filter groups with size greater than 2.

Practice

1. What does the groupBy function do in Kotlin?

easy

A. It sorts a list in ascending order.

B. It organizes a list into a map based on a key selector function.

C. It removes duplicate elements from a list.

D. It converts a list into a set.

Solution

1. Step 1: Understand the purpose of groupBy

   The groupBy function takes a list and groups its elements by a key, returning a map.

2. Step 2: Compare with other functions

   Sorting, removing duplicates, and converting to set do different things than grouping by a key.

3. Final Answer:

   It organizes a list into a map based on a key selector function. -> Option B

4. Quick Check:

   groupBy = organize by key [OK]

Hint: Remember: groupBy creates categories as map keys [OK]

Common Mistakes:

• Confusing groupBy with sorting functions
• Thinking groupBy removes duplicates
• Mixing groupBy with set conversion

2. Which of the following is the correct syntax to group a list of strings by their first character in Kotlin?

easy

A. val result = list.groupBy { it.first() }

B. val result = list.groupBy(it -> it.first())

C. val result = list.groupBy { it.first }

D. val result = list.groupBy { first(it) }

Solution

1. Step 1: Review Kotlin lambda syntax

   Kotlin uses curly braces with it for single-parameter lambdas: { it.first() }.

2. Step 2: Check each option

   val result = list.groupBy { it.first() } uses correct Kotlin syntax. val result = list.groupBy(it -> it.first()) uses arrow syntax which is invalid here. val result = list.groupBy { it.first } omits parentheses on first() method call, causing compile error. val result = list.groupBy { first(it) } calls undefined function first().

3. Final Answer:

   val result = list.groupBy { it.first() } -> Option A

4. Quick Check:

   Use curly braces and it.first() [OK]

Hint: Use curly braces and 'it' for lambda in groupBy [OK]

Common Mistakes:

• Using arrow syntax inside groupBy lambda
• Calling undefined functions like first(it)
• Omitting parentheses on first() method call

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

val numbers = listOf(1, 2, 3, 4, 5, 6)
val grouped = numbers.groupBy { if (it % 2 == 0) "even" else "odd" }
println(grouped)

medium

A. {odd=[1, 3, 5], even=[2, 4, 6]}

B. {1=[1], 2=[2], 3=[3], 4=[4], 5=[5], 6=[6]}

C. {true=[2, 4, 6], false=[1, 3, 5]}

D. SyntaxError

Solution

1. Step 1: Understand the grouping condition

   The lambda groups numbers by "even" if divisible by 2, else "odd".

2. Step 2: Group numbers accordingly

   Numbers 2,4,6 go to "even"; 1,3,5 go to "odd".

3. Final Answer:

   {odd=[1, 3, 5], even=[2, 4, 6]} -> Option A

4. Quick Check:

   Group by even/odd labels [OK]

Hint: Check grouping keys carefully for expected output [OK]

Common Mistakes:

• Confusing keys with values in output
• Expecting boolean keys instead of strings
• Thinking groupBy sorts the list

4. Identify the error in this Kotlin code snippet:

val words = listOf("apple", "banana", "apricot")
val grouped = words.groupBy { it[0] }
println(grouped)

medium

A. Indexing with it[0] is invalid for strings

B. Must convert characters to strings before grouping

C. groupBy requires a function, not a lambda

D. No error; code runs and groups by first character

Solution

1. Step 1: Check indexing on strings

   In Kotlin, strings support indexing with [0] to get first character.

2. Step 2: Validate groupBy usage

   The lambda { it[0] } is valid and groups words by their first character.

3. Final Answer:

   No error; code runs and groups by first character -> Option D

4. Quick Check:

   Strings support indexing; groupBy accepts lambdas [OK]

Hint: Remember strings can be indexed like arrays in Kotlin [OK]

Common Mistakes:

• Thinking indexing strings causes error
• Believing groupBy needs named functions only
• Assuming characters must be converted to strings

5. Given a list of people represented as pairs Pair<String, Int> where the first element is the name and the second is the age, how would you group them by age category "minor" (age < 18) or "adult" (age >= 18) using groupBy?

hard

A. people.groupBy { it.first }

B. people.groupBy { it.first.length }

C. people.groupBy { if (it.second < 18) "minor" else "adult" }

D. people.groupBy { it.second }

Solution

1. Step 1: Identify grouping key based on age

   We want to group by "minor" if age < 18, else "adult".

2. Step 2: Use groupBy with conditional key

   Access age with it.second and apply if-else to return category string.

3. Final Answer:

   people.groupBy { if (it.second < 18) "minor" else "adult" } -> Option C

4. Quick Check:

   Group by age category using conditional [OK]

Hint: Use if-else inside groupBy lambda for categories [OK]

Common Mistakes:

• Grouping by name or name length instead of age
• Grouping by exact age instead of categories
• Confusing first and second in Pair