Map, filter, reduce patterns in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of map and filter combination

What is the output of the following Swift code?

Swift

let numbers = [1, 2, 3, 4, 5]
let result = numbers.map { $0 * 2 }.filter { $0 > 5 }
print(result)

A [1, 2, 3, 4, 5]

B [2, 4]

C [6, 8, 10]

D [10]

💡 Hint

First multiply each number by 2, then keep only those greater than 5.

The map doubles each number: [2, 4, 6, 8, 10]. The filter keeps numbers > 5: [6, 8, 10].

❓ Predict Output

intermediate

Reduce to sum even numbers

What is the output of this Swift code?

Swift

let numbers = [1, 2, 3, 4, 5, 6]
let sumEven = numbers.filter { $0 % 2 == 0 }.reduce(0) { $0 + $1 }
print(sumEven)

A 21

B 12

C 0

D 6

💡 Hint

Filter even numbers, then add them all together.

Even numbers are [2, 4, 6]. Their sum is 2 + 4 + 6 = 12.

🔧 Debug

advanced

Identify the error in reduce usage

What error does this Swift code produce?

Swift

let words = ["apple", "banana", "cherry"]
let sentence = words.reduce { $0 + ", " + $1 }
print(sentence)

A Runtime error: index out of range

B Missing initial result in reduce causes a compile error

C Type mismatch error

D No error, prints "apple, banana, cherry"

💡 Hint

Reduce without an initial value uses the first element as the initial accumulator if the array is non-empty.

No error. Swift's `reduce` without an initial value uses the first element ("apple") as the starting accumulator and applies the closure to combine with subsequent elements: "apple, banana" then "apple, banana, cherry".

❓ Predict Output

advanced

Chained map, filter, reduce output

What is the output of this Swift code?

Swift

let values = [3, 6, 9, 12]
let result = values.map { $0 / 3 }.filter { $0 % 2 == 0 }.reduce(1) { $0 * $1 }
print(result)

A 8

B 24

C 6

D 1

💡 Hint

Divide each by 3, keep even numbers, then multiply all starting from 1.

Map: [1, 2, 3, 4]. Filter even: [2, 4]. Reduce multiply: 1 * 2 * 4 = 8.

🧠 Conceptual

expert

Understanding reduce initial value impact

Given the array [2, 4, 6], what is the value of 'result' after running this Swift code? let result = [2, 4, 6].reduce(10) { acc, val in acc - val }

A −2

B −18

C 12

D 10

💡 Hint

Start from 10, subtract each element in order.

Calculation: 10 - 2 = 8; 8 - 4 = 4; 4 - 6 = -2.

Practice

1. What does the map function do in Swift when applied to an array?

easy

A. It changes each item in the array based on a rule.

B. It removes items that don't match a condition.

C. It combines all items into a single value.

D. It sorts the array in ascending order.

Solution

1. Step 1: Understand the purpose of map

   The map function applies a transformation to each element in the array, creating a new array with the changed items.

2. Step 2: Compare with other functions

   Unlike filter which removes items, or reduce which combines items, map changes each item individually.

3. Final Answer:

   It changes each item in the array based on a rule. -> Option A

4. Quick Check:

   map = change each item [OK]

Hint: Map transforms every item, filter picks, reduce combines [OK]

Common Mistakes:

• Confusing map with filter or reduce
• Thinking map removes items
• Assuming map combines items into one

2. Which of the following is the correct syntax to use filter on an array numbers to keep only even numbers?

easy

A. let evens = numbers.map { $0 % 2 == 0 }

B. let evens = numbers.filter($0 % 2 == 0)

C. let evens = numbers.reduce(0) { $0 % 2 == 0 }

D. let evens = numbers.filter { $0 % 2 == 0 }

Solution

1. Step 1: Identify correct filter syntax

   The filter method uses a closure with a condition inside curly braces, like { $0 % 2 == 0 }, to keep items matching the condition.

2. Step 2: Check each option

   let evens = numbers.filter { $0 % 2 == 0 } correctly uses filter with the closure. let evens = numbers.map { $0 % 2 == 0 } uses map which changes items, not filters. let evens = numbers.reduce(0) { $0 % 2 == 0 } misuses reduce. let evens = numbers.filter($0 % 2 == 0) misses the closure braces.

3. Final Answer:

   let evens = numbers.filter { $0 % 2 == 0 } -> Option D

4. Quick Check:

   filter syntax = filter { condition } [OK]

Hint: Filter needs braces and a condition inside { } [OK]

Common Mistakes:

• Using map instead of filter
• Omitting braces in closure
• Using reduce incorrectly for filtering

3. What is the output of this Swift code?

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

medium

A. 0

B. [1, 2, 3, 4]

C. 10

D. Error

Solution

1. Step 1: Understand reduce with initial value 0

   The reduce starts with 0 and adds each number in the array: 0 + 1 + 2 + 3 + 4.

2. Step 2: Calculate the sum

   Adding all numbers: 0 + 1 = 1, 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10.

3. Final Answer:

   10 -> Option C

4. Quick Check:

   Sum of [1,2,3,4] = 10 [OK]

Hint: Reduce with + sums all items starting from initial value [OK]

Common Mistakes:

• Confusing reduce with map or filter output
• Forgetting initial value in reduce
• Expecting array output instead of single value

4. Identify the error in this Swift code snippet:

let words = ["apple", "banana", "cherry"]
let filtered = words.filter { word in
    word.count > 5
}
print(filtered)

medium

A. No error, code runs correctly

B. Incorrect use of filter instead of map

C. Using word.count instead of word.length

D. Missing return keyword inside filter closure

Solution

1. Step 1: Check filter closure syntax

   The closure uses implicit return because it is a single expression, so no explicit return is needed.

2. Step 2: Verify property usage

   word.count is the correct way to get string length in Swift, so no error there.

3. Final Answer:

   No error, code runs correctly -> Option A

4. Quick Check:

   Single-expression closures return implicitly [OK]

Hint: Single-expression closures return automatically, no return needed [OK]

Common Mistakes:

• Adding unnecessary return keyword
• Confusing count with length property
• Thinking filter needs map syntax

5. Given an array of integers, how would you use map, filter, and reduce together to find the sum of squares of only the odd numbers? Choose the correct Swift code.

hard

A. let sum = numbers.map { $0 * $0 }.filter { $0 % 2 != 0 }.reduce(0, +)

B. let sum = numbers.filter { $0 % 2 != 0 }.map { $0 * $0 }.reduce(0, +)

C. let sum = numbers.reduce(0) { $0 + $1 * $1 }

D. let sum = numbers.filter { $0 % 2 == 0 }.map { $0 * $0 }.reduce(0, +)

Solution

1. Step 1: Filter odd numbers first

   We want only odd numbers, so use filter { $0 % 2 != 0 } to keep them.

2. Step 2: Map to square each odd number

   Use map { $0 * $0 } to get squares of the filtered odd numbers.

3. Step 3: Reduce to sum all squares

   Use reduce(0, +) to add all squared values into one sum.

4. Final Answer:

   let sum = numbers.filter { $0 % 2 != 0 }.map { $0 * $0 }.reduce(0, +) -> Option B

5. Quick Check:

   Filter odds, map squares, reduce sum [OK]

Hint: Filter odds first, then map squares, finally reduce sum [OK]

Common Mistakes:

• Mapping before filtering odd numbers
• Filtering even numbers by mistake
• Using reduce without mapping squares