Sorted and sortedBy in Kotlin - Mini Project: Build & Apply

Sorting a List of Fruits by Name and Length

📖 Scenario: You work in a fruit store and want to organize your list of fruits. Sometimes you want to see them in alphabetical order, and other times by how long their names are.

🎯 Goal: Learn how to use sorted() to sort a list alphabetically and sortedBy() to sort by a specific property, like the length of the fruit names.

📋 What You'll Learn

Create a list of fruits with exact names

Create a variable to hold the sorted list alphabetically

Create a variable to hold the list sorted by length of fruit names

Print both sorted lists exactly as specified

💡 Why This Matters

🌍 Real World

Sorting lists is common in apps like shopping lists, contact lists, or any place where you want to organize items for easier viewing.

💼 Career

Understanding sorting helps in data processing, UI display, and backend logic in many software development roles.

1

Create the list of fruits

Create a list called fruits with these exact string values in this order: "Banana", "Apple", "Cherry", "Date", "Elderberry"

Kotlin

// Create the list of fruits
// Your code here

Hint

Use listOf() to create a list with the exact fruit names.

2

Create a variable for alphabetical sorting

Create a variable called sortedFruits that holds the list fruits sorted alphabetically using sorted()

Kotlin

val fruits = listOf("Banana", "Apple", "Cherry", "Date", "Elderberry")
// Create sortedFruits using sorted() to sort fruits alphabetically
// Your code here

Hint

Use sorted() on the fruits list to get an alphabetically sorted list.

3

Create a variable for sorting by length

Create a variable called sortedByLength that holds the list fruits sorted by the length of each fruit name using sortedBy and the lambda { it.length }

Kotlin

val fruits = listOf("Banana", "Apple", "Cherry", "Date", "Elderberry")
val sortedFruits = fruits.sorted()
// Create sortedByLength using sortedBy with it.length
// Your code here

Hint

Use sortedBy { it.length } to sort the list by the length of each fruit name.

4

Print both sorted lists

Print the variable sortedFruits on one line and then print sortedByLength on the next line exactly as lists (use println)

Kotlin

val fruits = listOf("Banana", "Apple", "Cherry", "Date", "Elderberry")
val sortedFruits = fruits.sorted()
val sortedByLength = fruits.sortedBy { it.length }
// Print sortedFruits and sortedByLength
// Your code here

Hint

Use println() to print each sorted list on its own line.

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