Sorted and sortedBy in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the list in ascending order.

Kotlin

val numbers = listOf(5, 3, 8, 1)
val sortedNumbers = numbers.[1]()

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

AorderBy

Bsort

Csorted

Darrange

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sort' which is not a function on immutable lists.

Using 'orderBy' or 'arrange' which are not Kotlin functions.

The sorted() function returns a new list with elements sorted in ascending order.

2fill in blank

medium

Complete the code to sort the list of strings by their length.

Kotlin

val words = listOf("apple", "banana", "kiwi", "pear")
val sortedWords = words.[1] { it.length }

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

AsortedBy

Bsorted

CsortBy

DorderBy

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sorted' which sorts by natural order, not by length.

Using 'sortBy' which is a mutable list function.

The sortedBy function sorts elements according to the value returned by the selector function, here it.length.

3fill in blank

hard

Fix the error in the code to sort the list of pairs by the second value.

Kotlin

val pairs = listOf(Pair("a", 3), Pair("b", 1), Pair("c", 2))
val sortedPairs = pairs.[1] { it.second }

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

Asort

BsortedBy

Csorted

DsortBy

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sortBy' on an immutable list causes an error.

Using 'sorted' without a selector sorts by natural order.

sortedBy returns a new sorted list by the selector. sortBy works only on mutable lists.

4fill in blank

hard

Fill both blanks to create a map of words to their lengths, including only words longer than 3 characters.

Kotlin

val words = listOf("cat", "elephant", "dog", "horse")
val wordLengths = { [1] : [2] for word in words if word.length > 3 }

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

Aword

Blen(word)

Cword.length

Dword.size

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'len(word)' which is Python syntax, not Kotlin.

Using 'word.size' which is not a property of String in Kotlin.

The map keys are the words themselves, and the values are their lengths using word.length.

5fill in blank

hard

Fill all three blanks to create a map of uppercase words to their lengths, including only words longer than 4 characters.

Kotlin

val words = listOf("apple", "bat", "carrot", "dog")
val result = { [1] : [2] for word in words if word.[3] 4 }

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

Aword.uppercase()

Bword.length

Clength

Dlength()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length()' which is not a Kotlin property but a method in some languages.

Using 'word.length()' which is invalid in Kotlin.

The key is the uppercase word using word.uppercase(), the value is the length with word.length, and the condition checks word.length > 4.

Practice

1. What does the sorted() function do in Kotlin?

easy

A. It sorts the original list in place and returns nothing.

B. It arranges elements randomly in a new list.

C. It returns a new list with elements arranged in natural ascending order.

D. It filters elements based on a condition.

Solution

1. Step 1: Understand the purpose of sorted()

   The sorted() function creates a new list where elements are arranged in natural ascending order (numbers from smallest to largest, strings alphabetically).

2. Step 2: Check if original list changes

   The original list remains unchanged because sorted() returns a new sorted list without modifying the original.

3. Final Answer:

   It returns a new list with elements arranged in natural ascending order. -> Option C

4. Quick Check:

   sorted() = new ascending list [OK]

Hint: sorted() always returns a new ascending list [OK]

Common Mistakes:

• Thinking sorted() changes the original list
• Confusing sorted() with filtering
• Assuming sorted() arranges elements randomly

2. Which of the following is the correct syntax to sort a list of strings by their length using sortedBy in Kotlin?

easy

A. list.sortedBy { it.length }

B. list.sortedBy.length()

C. list.sortedBy(it.length)

D. list.sortedBy { length(it) }

Solution

1. Step 1: Identify correct lambda syntax for sortedBy

   The sortedBy function takes a lambda expression that returns the property to sort by. The correct syntax is list.sortedBy { it.property }.

2. Step 2: Check each option

   list.sortedBy { it.length } uses { it.length } which correctly accesses the length property of each string. Other options have syntax errors or incorrect usage.

3. Final Answer:

   list.sortedBy { it.length } -> Option A

4. Quick Check:

   sortedBy uses lambda with it.property [OK]

Hint: Use curly braces and 'it' to access property in sortedBy [OK]

Common Mistakes:

• Using parentheses instead of curly braces
• Calling length() as a function incorrectly
• Passing property outside lambda

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

val numbers = listOf(5, 3, 8, 1)
val sortedNumbers = numbers.sortedBy { it % 3 }
println(sortedNumbers)

medium

A. [1, 3, 5, 8]

B. [3, 6, 9, 12]

C. [1, 5, 3, 8]

D. [3, 1, 5, 8]

Solution

1. Step 1: Calculate the sorting key for each element

   Calculate it % 3 for each number: 5 % 3 = 2, 3 % 3 = 0, 8 % 3 = 2, 1 % 3 = 1.

2. Step 2: Sort numbers by these keys

   Sort by keys in ascending order: 0 (3), 1 (1), 2 (5), 2 (8). So sorted list is [3, 1, 5, 8].

3. Final Answer:

   [3, 1, 5, 8] -> Option D

4. Quick Check:

   sortedBy { it % 3 } = [3, 1, 5, 8] [OK]

Hint: Calculate sorting key for each element before sorting [OK]

Common Mistakes:

• Sorting by values instead of keys
• Confusing modulo results
• Assuming sortedBy sorts original list

4. Identify the error in this Kotlin code snippet:

val fruits = listOf("apple", "banana", "cherry")
val sortedFruits = fruits.sortedBy(it.length)
println(sortedFruits)

medium

A. sortedBy cannot be used on lists of strings.

B. Missing curly braces around the lambda expression in sortedBy.

C. The list should be mutable to use sortedBy.

D. The println statement is incorrect.

Solution

1. Step 1: Check syntax of sortedBy usage

   The sortedBy function requires a lambda expression inside curly braces, like { it.length }. The code uses parentheses without braces, which is incorrect.

2. Step 2: Verify other options

   sortedBy works on lists of strings, the list can be immutable, and println is correct. So only the lambda syntax is wrong.

3. Final Answer:

   Missing curly braces around the lambda expression in sortedBy. -> Option B

4. Quick Check:

   sortedBy needs { } for lambda [OK]

Hint: Always use curly braces for lambda in sortedBy [OK]

Common Mistakes:

• Using parentheses instead of curly braces for lambda
• Thinking sortedBy needs mutable list
• Misunderstanding println usage

5. Given a list of data class objects data class Person(val name: String, val age: Int), how would you sort the list first by age ascending, then by name alphabetically using Kotlin's sorting functions?

hard

A. persons.sortedWith(compareBy({ it.age }, { it.name }))

B. persons.sortedBy { it.age + it.name }

C. persons.sortedBy { it.name }.sortedBy { it.age }

D. persons.sortedByDescending { it.age }.sortedBy { it.name }

Solution

1. Step 1: Understand multi-level sorting in Kotlin

   To sort by multiple properties, use sortedWith(compareBy(...)) passing lambdas for each property in priority order.

2. Step 2: Analyze options

   persons.sortedWith(compareBy({ it.age }, { it.name })) correctly uses compareBy with age first, then name. persons.sortedBy { it.age + it.name } incorrectly adds age and name (different types). persons.sortedBy { it.name }.sortedBy { it.age } sorts twice but second sort overrides first. persons.sortedByDescending { it.age }.sortedBy { it.name } sorts age descending, not ascending.

3. Final Answer:

   persons.sortedWith(compareBy({ it.age }, { it.name })) -> Option A

4. Quick Check:

   Use sortedWith(compareBy(...)) for multi-level sort [OK]

Hint: Use sortedWith(compareBy(...)) for sorting by multiple keys [OK]

Common Mistakes:

• Adding different types in sortedBy
• Chaining sortedBy loses previous sort
• Using sortedByDescending when ascending needed