Collections (List, Map, Set) in Android Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ ui_behavior

intermediate

What is the output of this Kotlin code using a List?

Consider this Kotlin code snippet that creates and prints a list. What will be printed?

Android Kotlin

val fruits = listOf("Apple", "Banana", "Cherry")
println(fruits[1])

A Cherry

B Apple

C Banana

D IndexOutOfBoundsException

💡 Hint

Remember that list indices start at zero.

In Kotlin, list indices start at 0, so fruits[1] accesses the second item, which is "Banana".

📝 Syntax

intermediate

What error does this Kotlin Map code produce?

Examine this Kotlin code that tries to create a map. What error will it cause?

Android Kotlin

val map = mapOf("one" to 1, "two" 2, "three" to 3)

A No error, map created successfully

B TypeError: Cannot assign Int to String

C RuntimeException: Duplicate keys

D SyntaxError: Missing 'to' keyword

💡 Hint

Check the syntax for key-value pairs in mapOf.

In Kotlin, mapOf requires key-value pairs separated by 'to'. The pair "two" 2 is missing 'to', causing a syntax error.

❓ lifecycle

advanced

What happens when you add duplicate elements to a Kotlin Set?

Given this Kotlin code, what will be the size of the set after adding duplicates?

Android Kotlin

val numbers = mutableSetOf(1, 2, 3)
numbers.add(2)
numbers.add(4)
println(numbers.size)

A 5

B 4

C 3

D RuntimeException

💡 Hint

Sets do not allow duplicate elements.

Sets only keep unique elements. Adding 2 again does not increase size. Adding 4 adds a new element. So size is 4.

🔧 Debug

advanced

Why does this Kotlin code throw an exception?

This Kotlin code tries to access a map value. Why does it throw an exception?

Android Kotlin

val map = mapOf("a" to 1, "b" to 2)
val value = map["c"]!!
println(value)

A NullPointerException because key "c" is missing and !! forces non-null

B No exception, prints 0

C KeyNotFoundException because "c" is not in map

D Compilation error due to !! operator

💡 Hint

What does the !! operator do when the value is null?

The !! operator asserts non-null. Since map["c"] returns null, !! throws NullPointerException.

🧠 Conceptual

expert

How many items are in this Kotlin collection after these operations?

What is the size of the collection after this Kotlin code runs?

Android Kotlin

val list = listOf(1, 2, 3, 4)
val set = list.toMutableSet()
set.remove(2)
set.add(5)
val map = set.associateWith { it * 2 }
println(map.size)

A 4

B 5

C 3

D 6

💡 Hint

Track each step: list to set, remove, add, then map size.

Starting with list of 4 items, converted to set (4 unique items). Remove one (2), add one (5), set size remains 4. Map keys come from set, so map size is 4.

Practice

1. Which Kotlin collection type should you use to store a list of names where order matters and duplicates are allowed?

easy

A. List

B. Map

C. Set

D. Array

Solution

1. Step 1: Understand collection types

   List keeps items in order and allows duplicates, Map stores key-value pairs, Set stores unique items without duplicates.

2. Step 2: Match requirement to collection

   Since order matters and duplicates are allowed, List fits best.

3. Final Answer:

   List -> Option A

4. Quick Check:

   Order + duplicates = List [OK]

Hint: Remember: List = ordered + duplicates allowed [OK]

Common Mistakes:

• Choosing Set which disallows duplicates
• Choosing Map which stores key-value pairs
• Confusing Array with List in Kotlin

2. Which of the following is the correct syntax to create a mutable map in Kotlin?

easy

A. val map = mutableListOf()

B. val map = mapOf()

C. val map = mutableMapOf()

D. val map = setOf()

Solution

1. Step 1: Identify mutable map syntax

   mutableMapOf creates a mutable map; mapOf creates an immutable map.

2. Step 2: Check other options

   mutableListOf creates a list, setOf creates a set; both are incorrect for map.

3. Final Answer:

   val map = mutableMapOf() -> Option C

4. Quick Check:

   Mutable map uses mutableMapOf [OK]

Hint: Mutable map uses mutableMapOf() function [OK]

Common Mistakes:

• Using mapOf which is immutable
• Using mutableListOf instead of mutableMapOf
• Using setOf which creates a set, not a map

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

val set = setOf(1, 2, 2, 3)
println(set.size)

medium

A. 3

B. 2

C. 4

D. Error

Solution

1. Step 1: Understand set behavior

   Set stores unique elements only, so duplicates are removed.

2. Step 2: Count unique elements

   Elements are 1, 2, 3 so size is 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Set removes duplicates, size = 3 [OK]

Hint: Set size counts unique elements only [OK]

Common Mistakes:

• Counting duplicates in set size
• Assuming set keeps duplicates
• Expecting runtime error for duplicates

4. Identify the error in this Kotlin code snippet:

val map = mapOf("a" to 1, "b" to 2)
map["c"] = 3

medium

A. map["c"] = 3 is correct syntax

B. Syntax error in map declaration

C. Key "c" already exists

D. mapOf creates an immutable map, cannot add elements

Solution

1. Step 1: Check mapOf mutability

   mapOf creates an immutable map, so you cannot add or change elements.

2. Step 2: Analyze assignment

   map["c"] = 3 tries to add a new key-value pair, which is not allowed on immutable map.

3. Final Answer:

   mapOf creates an immutable map, cannot add elements -> Option D

4. Quick Check:

   mapOf = immutable map, no additions allowed [OK]

Hint: mapOf is immutable; use mutableMapOf to add items [OK]

Common Mistakes:

• Assuming mapOf is mutable
• Thinking syntax is wrong
• Ignoring immutability of mapOf

5. Given a list of words, how can you create a map that counts how many times each word appears? Choose the correct Kotlin code snippet.

hard

A. val counts = words.filter { it.length > 1 }

B. val counts = words.groupingBy { it }.eachCount()

C. val counts = words.distinct()

D. val counts = words.toSet().map { it.length }

Solution

1. Step 1: Understand the goal

   We want a map with words as keys and their counts as values.

2. Step 2: Analyze options

   val counts = words.groupingBy { it }.eachCount() uses groupingBy and eachCount to count occurrences correctly. Others do not count frequencies.

3. Final Answer:

   val counts = words.groupingBy { it }.eachCount() -> Option B

4. Quick Check:

   groupingBy + eachCount counts items [OK]

Hint: Use groupingBy and eachCount to count list items [OK]

Common Mistakes:

• Using toSet which removes duplicates but doesn't count
• Using map to get lengths instead of counts
• Filtering instead of counting