Range operators (... and ..<) in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a closed range from 1 to 5.

Swift

let range = 1 [1] 5
print(Array(range))

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

A..<

B...

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using ..< excludes the last number.

Using - or + is not valid for ranges.

The ... operator creates a closed range including both start and end values.

2fill in blank

medium

Complete the code to create a half-open range from 1 up to but not including 5.

Swift

let range = 1 [1] 5
print(Array(range))

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

A..<

B...

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using ... includes the last number.

Using + or - is not valid for ranges.

The ..< operator creates a half-open range that excludes the upper bound.

3fill in blank

hard

Fix the error in the code to print numbers from 1 to 4 using a half-open range.

Swift

for i in 1 [1] 5 {
    print(i)
}

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

A...

B+

C..<

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using ... includes 5, which is not desired here.

Using + or - is invalid syntax.

Using ..< creates a range from 1 up to but not including 5, so it prints 1 to 4.

4fill in blank

hard

Fill both blanks to create a dictionary with word lengths for words longer than 3 characters.

Swift

let words = ["apple", "bat", "carrot", "dog"]
let lengths = Dictionary(uniqueKeysWithValues: words.filter { word in [2] }.map { word in (word, [1]) } )

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

Aword.count

Bword.count > 3

Cword.count < 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using word instead of word.count for length.

Filtering words with length less than 3.

We use word.count to get the length and filter words with length greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase words and their lengths for words longer than 3 characters.

Swift

let words = ["apple", "bat", "carrot", "dog"]
let lengths = Dictionary(uniqueKeysWithValues: words.filter { word in [3] }.map { word in ([1], [2]) } )

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

Aword.uppercased()

Bword.count

Cword.count > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using word instead of word.uppercased() for keys.

Not filtering words by length.

The keys are uppercase words, values are their lengths, filtered by length greater than 3.