GroupBy for categorization in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Kotlin code using groupBy?

Consider this Kotlin code that groups a list of words by their first letter. What will be printed?

Kotlin

val words = listOf("apple", "banana", "apricot", "blueberry", "cherry")
val grouped = words.groupBy { it.first() }
println(grouped)

A [[apple, apricot], [banana, blueberry], [cherry]]

B {apple=[a, p, p, l, e], banana=[b, a, n, a, n, a], apricot=[a, p, r, i, c, o, t], blueberry=[b, l, u, e, b, e, r, r, y], cherry=[c, h, e, r, r, y]}

C {a=[apple, apricot], b=[banana, blueberry], c=[cherry]}

D {a=[apple], b=[banana], a=[apricot], b=[blueberry], c=[cherry]}

💡 Hint

Remember that groupBy creates a map where keys are the grouping criteria and values are lists of items matching that key.

The groupBy function groups words by their first character. So all words starting with 'a' are grouped together, and so on.

❓ Predict Output

intermediate

What is the output of grouping numbers by even or odd?

Given this Kotlin code, what will be printed?

Kotlin

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

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

B [[1, 3, 5], [2, 4, 6]]

C {0=[2, 4, 6], 1=[1, 3, 5]}

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

💡 Hint

Look at the string keys used in the groupBy lambda.

The lambda returns "even" or "odd" strings, so the map keys are those strings with lists of numbers accordingly.

🔧 Debug

advanced

Why does this Kotlin code cause a compilation error?

Look at this Kotlin code that tries to group a list of strings by length but causes a compilation error. What is the cause?

Kotlin

val words = listOf("cat", "dog", "bird")
val grouped = words.groupBy { length }
println(grouped)

A The lambda uses 'length' without 'it.', so 'length' is undefined.

B The list 'words' is immutable, so groupBy cannot be called.

C The groupBy function requires a return type of Int, but 'length' is a String.

D The println statement is missing parentheses.

💡 Hint

Check how to access the length property of each string inside the lambda.

Inside the lambda, 'it' refers to each string. You must write 'it.length' to access the length property. Writing just 'length' causes an error because 'length' is not defined in that scope.

🧠 Conceptual

advanced

How many groups are created by this Kotlin groupBy call?

Given this code, how many keys will the resulting map have?

Kotlin

val items = listOf("red", "green", "blue", "yellow", "black")
val grouped = items.groupBy { it.length }
println(grouped.keys.size)

A 5

B 4

C 3

D 2

💡 Hint

Count the distinct lengths of the strings in the list.

The lengths are: red(3), green(5), blue(4), yellow(6), black(5). Distinct lengths are 3, 4, 5, 6, so 4 groups.

❓ Predict Output

expert

What is the output of this Kotlin code grouping by first letter and counting?

What will this Kotlin code print?

Kotlin

val words = listOf("ant", "ape", "bat", "bear", "cat", "car")
val grouped = words.groupBy { it.first() }.mapValues { it.value.size }
println(grouped)

A {a=2, b=2, c=2}

B {a=2, b=3, c=2}

C {a=3, b=2, c=2}

D {a=[ant, ape], b=[bat, bear], c=[cat, car]}

💡 Hint

Look at how mapValues transforms the grouped map.

The groupBy groups words by first letter, then mapValues replaces each list with its size. Each letter has exactly 2 words, so the counts are 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