For loop with index (withIndex) in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to loop over the list with index and print each index.

Kotlin

val fruits = listOf("Apple", "Banana", "Cherry")
for (([1], fruit) in fruits.withIndex()) {
    println("Index: $[1]")
}

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

Apos

Bi

Cindex

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that does not match the index variable in the loop.

Forgetting to use parentheses around the destructured variables.

The variable 'index' is commonly used to represent the index in a for loop with withIndex().

2fill in blank

medium

Complete the code to print both index and fruit name inside the loop.

Kotlin

val fruits = listOf("Apple", "Banana", "Cherry")
for ((index, [1]) in fruits.withIndex()) {
    println("$index: $[1]")
}

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

Aitem

Bfruit

Celement

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not descriptive of the list elements.

Mixing up the order of index and element in the destructuring.

The variable 'fruit' is used to represent each element in the list during iteration.

3fill in blank

hard

Fix the error in the loop variable declaration to correctly destructure index and element.

Kotlin

val numbers = listOf(10, 20, 30)
for ([1], num in numbers.withIndex()) {
    println("$num at index $[1]")
}

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

A(index, num)

Bindex

C(index,)

D(index)

Attempts:

3 left

💡 Hint

Common Mistakes

Not using parentheses for destructuring.

Only declaring one variable instead of two.

The correct syntax for destructuring both index and element is '(index, num)'.

4fill in blank

hard

Fill both blanks to create a map of index to fruit name for fruits longer than 5 characters.

Kotlin

val fruits = listOf("Apple", "Banana", "Cherry", "Date")
val longFruits = fruits.withIndex().filter { it.value.length [1] 5 }.associate { it.index to it.[2] }

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

A>

B<

Cvalue

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong comparison operator.

Using 'length' instead of 'value' in the associate function.

We filter fruits with length greater than 5 and associate index to the fruit value.

5fill in blank

hard

Fill all three blanks to create a list of strings showing index and fruit in uppercase for fruits starting with 'B' or later.

Kotlin

val fruits = listOf("Apple", "Banana", "Cherry", "Date")
val result = fruits.withIndex().filter { it.value[0] [1] 'B' }.map { "$[2]: $[3]" }
println(result)

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

A>=

Bit.index

Cit.value.uppercase()

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>=' in the filter.

Not converting fruit names to uppercase.

We filter fruits starting with letters 'B' or later, then map to strings with index and uppercase fruit name.

Practice

1. What does the withIndex() function do in a Kotlin for loop?

easy

A. It filters elements based on their index.

B. It sorts the list before looping.

C. It provides both the index and the value of each element during iteration.

D. It reverses the list before looping.

Solution

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

   The withIndex() function returns an iterable of pairs containing the index and the value of each element in the collection.

2. Step 2: Identify what the loop receives

   Using withIndex() in a for loop allows access to both the position (index) and the element (value) simultaneously.

3. Final Answer:

   It provides both the index and the value of each element during iteration. -> Option C

4. Quick Check:

   withIndex() = index + value [OK]

Hint: Remember: withIndex() gives index and value together [OK]

Common Mistakes:

• Thinking it sorts or filters the list
• Assuming it reverses the collection
• Confusing index with value only

2. Which of the following is the correct syntax to loop over a list items with index in Kotlin?

easy

A. for ((index, value) in items.withIndex()) { /* code */ }

B. for (index, value in items) { /* code */ }

C. for (index; value in items.withIndex()) { /* code */ }

D. for index, value in items.withIndex() { /* code */ }

Solution

1. Step 1: Recall Kotlin destructuring syntax in loops

   Kotlin allows destructuring pairs using parentheses in the for loop header, like for ((index, value) in ...).

2. Step 2: Match syntax with withIndex()

   The correct syntax uses items.withIndex() and destructures the pair into index and value.

3. Final Answer:

   for ((index, value) in items.withIndex()) { /* code */ } -> Option A

4. Quick Check:

   Destructure pairs with (index, value) [OK]

Hint: Use parentheses to destructure pairs in for loop [OK]

Common Mistakes:

• Omitting parentheses around (index, value)
• Using semicolon instead of comma
• Missing 'in' keyword

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

val data = listOf("apple", "banana", "cherry")
for ((i, v) in data.withIndex()) {
    println("$i: $v")
}

medium

A. 0: apple
1: banana
2: cherry

B. apple: 0
banana: 1
cherry: 2

C. 1: apple
2: banana
3: cherry

D. Error: cannot destructure

Solution

1. Step 1: Understand the list and loop

   The list contains three fruits: "apple", "banana", "cherry". The loop uses withIndex() to get index and value.

2. Step 2: Trace the loop output

   Each iteration prints the index and value separated by colon and space. Indices start at 0.

3. Final Answer:

   0: apple 1: banana 2: cherry -> Option A

4. Quick Check:

   Index starts at 0, prints index: value [OK]

Hint: Index starts at 0 in withIndex() loops [OK]

Common Mistakes:

• Assuming index starts at 1
• Swapping index and value in output
• Expecting error from destructuring

4. Identify the error in this Kotlin code snippet:

val numbers = arrayOf(10, 20, 30)
for (i, num in numbers.withIndex()) {
    println("$i -> $num")
}

medium

A. Cannot use withIndex() on arrays.

B. Missing parentheses around (i, num) in the for loop.

C. The println statement syntax is incorrect.

D. Variable names i and num are reserved keywords.

Solution

1. Step 1: Check for correct destructuring syntax

   In Kotlin, destructuring pairs in a for loop requires parentheses: for ((i, num) in ...).

2. Step 2: Verify other parts of the code

   Using withIndex() on arrays is valid. The println syntax and variable names are correct.

3. Final Answer:

   Missing parentheses around (i, num) in the for loop. -> Option B

4. Quick Check:

   Destructure pairs with parentheses [OK]

Hint: Always use parentheses to destructure pairs in for loops [OK]

Common Mistakes:

• Omitting parentheses in destructuring
• Thinking withIndex() doesn't work on arrays
• Misreading println syntax

5. Given a list of database table names tables = listOf("users", "orders", "products"), how can you create a map of index to table name using a Kotlin for loop with withIndex()?

hard

A. val map = mutableMapOf()
for (t in tables) {
map[t] = t.length
}

B. val map = mutableMapOf()
for ((i, t) in tables.withIndex()) {
map[t] = i
}

C. val map = mutableMapOf()
for (i in tables.withIndex()) {
map[i] = tables[i]
}

D. val map = mutableMapOf()
for ((i, t) in tables.withIndex()) {
map[i] = t
}

Solution

1. Step 1: Understand the goal

   We want a map where keys are indices (Int) and values are table names (String).

2. Step 2: Use withIndex() to get index and value

   Loop over tables.withIndex() and assign map[i] = t to build the map correctly.

3. Final Answer:

   val map = mutableMapOf() for ((i, t) in tables.withIndex()) { map[i] = t } -> Option D

4. Quick Check:

   Index to value mapping with withIndex() [OK]

Hint: Use withIndex() and map[i] = value to build index-value map [OK]

Common Mistakes:

• Swapping key and value types in map
• Not destructuring with parentheses
• Using wrong loop variable types