Lazy collections for performance in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this lazy map operation?

Consider the following Swift code using a lazy collection. What will be printed?

Swift

let numbers = [1, 2, 3, 4, 5]
let lazySquares = numbers.lazy.map { num in
    print("Mapping \(num)")
    return num * num
}
print("Before accessing elements")
let firstSquare = lazySquares.first
print("First square: \(firstSquare ?? 0)")

A Before accessing elements\nMapping 1\nFirst square: 1

B Mapping 1\nBefore accessing elements\nFirst square: 1

C Before accessing elements\nMapping 1\nMapping 2\nMapping 3\nMapping 4\nMapping 5\nFirst square: 1

D Mapping 1\nMapping 2\nMapping 3\nMapping 4\nMapping 5\nBefore accessing elements\nFirst square: 1

💡 Hint

Remember that lazy collections delay computation until elements are accessed.

The lazy map does not run the closure until the element is accessed. The print before accessing elements runs first. Then only the first element is mapped and printed.

❓ Predict Output

intermediate

How many elements are processed in this lazy filter?

Given this Swift code, how many times will the print statement inside the filter closure run?

Swift

let numbers = Array(1...10)
let lazyFiltered = numbers.lazy.filter { num in
    print("Filtering \(num)")
    return num % 2 == 0
}
let firstTwo = Array(lazyFiltered.prefix(2))

A 4 times

B 2 times

C 10 times

D 5 times

💡 Hint

The filter stops after finding enough elements for prefix(2).

The filter runs until it finds 2 even numbers. It checks numbers 1 to 4, printing 4 times. Then it stops.

🔧 Debug

advanced

Why does this lazy map not print anything?

Look at this Swift code using lazy map. Why does it produce no output?

Swift

let numbers = [1, 2, 3]
let lazyMapped = numbers.lazy.map { num in
    print("Mapping \(num)")
    return num * 2
}
// No further code

A Because the print statement is inside a closure that is immediately executed

B Because the lazy map closure is never executed without accessing elements

C Because the array is empty

D Because lazy collections do not support print statements

💡 Hint

Think about when lazy collections run their closures.

Lazy collections delay running the closure until elements are accessed. Since no element is accessed, the closure never runs.

❓ Predict Output

advanced

What is the output of this chained lazy operation?

What will this Swift code print?

Swift

let numbers = [1, 2, 3, 4, 5]
let lazyResult = numbers.lazy.filter { num in
    print("Filtering \(num)")
    return num % 2 != 0
}.map { num in
    print("Mapping \(num)")
    return num * 10
}
print("Start accessing")
let result = Array(lazyResult.prefix(2))
print("Result: \(result)")

A Start accessing\nFiltering 1\nFiltering 2\nFiltering 3\nFiltering 4\nFiltering 5\nMapping 1\nMapping 3\nMapping 5\nResult: [10, 30]

B Filtering 1\nFiltering 2\nFiltering 3\nFiltering 4\nFiltering 5\nMapping 1\nMapping 3\nMapping 5\nStart accessing\nResult: [10, 30]

C Start accessing\nFiltering 1\nMapping 1\nFiltering 2\nFiltering 3\nMapping 3\nResult: [10, 30]

D Start accessing\nFiltering 1\nMapping 1\nFiltering 2\nMapping 2\nFiltering 3\nMapping 3\nResult: [10, 20, 30]

💡 Hint

Remember lazy evaluation stops once prefix(2) elements are found.

The filter runs until it finds 2 odd numbers (1 and 3). It filters 1, 2, 3 and maps only those odd numbers. It stops after prefix(2).

🧠 Conceptual

expert

Why use lazy collections for performance in Swift?

Which of the following best explains why lazy collections improve performance?

A They automatically parallelize operations across CPU cores

B They compute all elements immediately to avoid delays later

C They store all intermediate results in memory for faster access

D They delay computation until elements are needed, reducing unnecessary work

💡 Hint

Think about when lazy collections run their operations.

Lazy collections delay running operations until elements are accessed, so they avoid doing work on elements that might never be used.