Map, filter, reduce patterns in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new array with each number doubled using map.

Swift

let numbers = [1, 2, 3, 4]
let doubled = numbers.[1] { $0 * 2 }

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Afilter

Breduce

Cmap

DcompactMap

Attempts:

3 left

💡 Hint

Common Mistakes

Using filter instead of map, which only selects elements but does not transform them.

The map function applies the given closure to each element and returns a new array with the results.

2fill in blank

medium

Complete the code to filter out numbers less than 3.

Swift

let numbers = [1, 2, 3, 4, 5]
let filtered = numbers.[1] { $0 >= 3 }

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Afilter

Breduce

Cmap

DflatMap

Attempts:

3 left

💡 Hint

Common Mistakes

Using map instead of filter, which transforms elements but does not remove any.

The filter function returns a new array containing only elements that satisfy the condition.

3fill in blank

hard

Fix the error in the code to sum all numbers using reduce.

Swift

let numbers = [1, 2, 3, 4]
let sum = numbers.reduce([1]) { $0 + $1 }

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A1

B0

C[]

D""

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as initial value, which adds an extra 1 to the sum.

Using an empty array or string, which causes a type error.

The initial value for summing numbers should be 0, so the sum starts from zero.

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths as values, filtering words longer than 3 characters.

Swift

let words = ["apple", "bat", "car", "door"]
let lengths = Dictionary(uniqueKeysWithValues: words.filter { $0.count [1] 3 }.map { [2] in ([2], [2].count) })

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A>

B<

Cword

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using 'count' instead of 'word' as the key in the tuple.

We filter words with length greater than 3, then map each word to a tuple of (word, length).

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase words as keys, their counts as values, filtering words with count greater than 1.

Swift

let words = ["apple", "bat", "apple", "car", "bat", "bat"]
let counts = words.reduce(into: [:]) { counts, [1] in
    counts[[2]] = (counts[[3]] ?? 0) + 1
}
let filteredCounts = counts.filter { $0.value > 1 }

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Aword

Bword.uppercased()

Attempts:

3 left

💡 Hint

Common Mistakes

Using the lowercase word as key instead of uppercase.

Using inconsistent keys for reading and writing counts.

We count occurrences of uppercase words. The key in the dictionary is the uppercase word, so we use word.uppercased() for keys and lookups.