For loop with index (withIndex) in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of for loop with withIndex on a list

What is the output of this Kotlin code snippet?

val fruits = listOf("apple", "banana", "cherry")
for ((index, fruit) in fruits.withIndex()) {
    println("$index: $fruit")
}

Kotlin

val fruits = listOf("apple", "banana", "cherry")
for ((index, fruit) in fruits.withIndex()) {
    println("$index: $fruit")
}

A

apple: 0
banana: 1
cherry: 2

B

0: apple
1: banana
2: cherry

C

apple
banana
cherry

D

0: apple
1: banana
2: cherry
3: null

💡 Hint

Remember that withIndex() provides both the index and the element in the loop.

The withIndex() function returns an iterable of IndexedValue objects, each containing an index and the corresponding element. The loop destructures these into index and fruit, printing them in order.

📝 Syntax

intermediate

Identify the syntax error in for loop with withIndex

Which option contains a syntax error in using withIndex() in a for loop?

Kotlin

val numbers = listOf(10, 20, 30)
for ((i, num) in numbers.withIndex()) {
    println("Index $i has value $num")
}

A for ((i, num) in numbers.withIndex()) { println(i + num) }

B } )mun + i(nltnirp { ))(xednIhtiw.srebmun ni )mun ,i(( rof

C for ((i, num) in numbers.withIndex()) println(i + num)

D for (i, num in numbers.withIndex()) { println(i + num) }

💡 Hint

Check how destructuring declarations are used in Kotlin for loops.

Option D misses parentheses around the destructuring variables, which is required syntax in Kotlin for destructuring in for loops.

❓ optimization

advanced

Optimizing iteration with withIndex on large lists

Given a large list, which approach is more efficient to print index and value pairs?
Options:
A) Using withIndex() in a for loop
B) Using indices property and accessing elements by index
C) Using forEachIndexed function
D) Using a while loop with manual index increment

A for ((index, value) in list.withIndex()) { println("$index: $value") }

B for (i in list.indices) { println("$i: ${list[i]}") }

C list.forEachIndexed { index, value -> println("$index: $value") }

D

var i = 0
while (i < list.size) { println("$i: ${list[i]}"); i++ }

💡 Hint

Consider readability and internal optimizations of Kotlin standard library functions.

forEachIndexed is optimized and concise, often preferred for readability and performance over manual loops or withIndex in large collections.

🔧 Debug

advanced

Debug the unexpected output in for loop with withIndex

What causes this Kotlin code to print incorrect indices?

val items = listOf("a", "b", "c")
for ((index, item) in items.withIndex()) {
    if (item == "b") continue
    println("$index: $item")
}

A The continue statement skips printing but does not skip index increment, so indices remain correct.

B The loop should use break instead of continue to avoid index issues.

C The index variable is not updated correctly because of continue, causing wrong indices.

D Using continue inside for loop with withIndex causes index to reset to 0.

💡 Hint

Think about how continue affects loop iteration and index values.

The continue statement skips the current iteration's print but the loop still increments index correctly. Indices remain consistent.

🧠 Conceptual

expert

Understanding withIndex behavior on different collections

Which statement about withIndex() in Kotlin is TRUE?

A withIndex() can be used on any Iterable and returns an IndexedValue sequence with index starting at 0.

B withIndex() modifies the original collection by adding index properties to elements.

C withIndex() returns indices and values only for List collections, not for Sets.

D withIndex() returns a Map of index to element for any collection.

💡 Hint

Recall what types of collections implement Iterable and how withIndex works.

withIndex() works on any Iterable, producing a sequence of IndexedValue objects with zero-based indices without modifying the original collection.

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