Zip for combining sequences in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift code using zip?

Consider the following Swift code that uses zip to combine two arrays. What will be printed?

Swift

let numbers = [1, 2, 3]
let words = ["one", "two", "three"]
for (number, word) in zip(numbers, words) {
    print("\(number): \(word)")
}

A one: 1\ntwo: 2\nthree: 3

B 1: one\n2: two

C 3: three\n2: two\n1: one

D 1: one\n2: two\n3: three

💡 Hint

Remember that zip pairs elements from both sequences in order.

The zip function pairs elements from numbers and words by their positions. So the first pair is (1, "one"), second is (2, "two"), and third is (3, "three"). The loop prints each pair in order.

❓ Predict Output

intermediate

What happens when zipping sequences of different lengths?

What will be the output of this Swift code?

Swift

let a = [10, 20, 30, 40]
let b = ["a", "b"]
for (num, letter) in zip(a, b) {
    print("\(num)\(letter)")
}

A 10a\n20b

B 10a\n20b\n30c\n40d

C 30c\n40d

D 10a\n20b\n30\n40

💡 Hint

Think about how zip handles sequences of different lengths.

The zip function stops when the shortest sequence ends. Here, b has only 2 elements, so only two pairs are formed.

🔧 Debug

advanced

Why does this Swift code using zip cause a runtime error?

Examine the code below. Why does it cause a runtime error?

Swift

let nums = [1, 2, 3]
let words = ["one", "two"]
let zipped = zip(nums, words)
let array = Array(zipped)
print(array[2])

A Type error because zipped is not convertible to Array

B Syntax error because zip requires sequences of equal length

C Index out of range error at runtime because zipped has only 2 elements

D No error, prints (3, "three")

💡 Hint

Check how many elements zipped contains and what index is accessed.

The zip pairs only up to the shortest sequence length, so zipped has 2 elements. Accessing array[2] causes an index out of range runtime error.

🧠 Conceptual

advanced

What is the type of the result of zip in Swift?

Given two arrays a and b, what is the type of zip(a, b) in Swift?

A Sequence<(Int, String)>

B Zip2Sequence<Array<Int>, Array<String>>

C [(Int, String)]

D Array<(Int, String)>

💡 Hint

Think about what zip returns before converting to an array.

The zip function returns a Zip2Sequence type that lazily pairs elements from the two sequences. It is not an array until converted.

🚀 Application

expert

How to sum pairs of numbers using zip in Swift?

You have two arrays of integers: arr1 and arr2. You want to create a new array where each element is the sum of the corresponding elements from arr1 and arr2. Which code snippet correctly does this using zip?

A let sums = zip(arr1, arr2).map { (a, b) in a + b }

B let sums = zip(arr1, arr2).map { $0 + $1 }

C let sums = zip(arr1, arr2).map { $0 * $1 }

D let sums = arr1.enumerated().map { $0.element + arr2[$0.offset] }

💡 Hint

Remember the closure syntax for map when using zip.

Option A correctly uses zip to pair elements and a closure with two parameters to sum them. Option A is incorrect because $0 is a tuple, so $0 + $1 is invalid. Option A works but does not use zip. Option A multiplies instead of sums.

Practice

1. What does the zip function do in Swift?

easy

A. Filters elements from a sequence based on a condition

B. Adds all elements of two sequences together

C. Sorts two sequences in ascending order

D. Combines two sequences into pairs until the shortest ends

Solution

1. Step 1: Understand the purpose of zip

   The zip function pairs elements from two sequences together, creating tuples.

2. Step 2: Recognize stopping condition

   It stops pairing when the shortest sequence runs out of elements.

3. Final Answer:

   Combines two sequences into pairs until the shortest ends -> Option D

4. Quick Check:

   Zip pairs sequences = D [OK]

Hint: Remember: zip pairs elements, stops at shortest sequence [OK]

Common Mistakes:

• Thinking zip adds or merges elements
• Assuming zip continues to longest sequence length
• Confusing zip with sorting or filtering

2. Which of the following is the correct syntax to zip two arrays a and b in Swift?

easy

A. let zipped = zip(a, b)

B. let zipped = zip[a, b]

C. let zipped = Array.zip(a, b)

D. let zipped = zip a, b

Solution

1. Step 1: Recall Swift function call syntax

   Functions are called with parentheses and arguments inside, like zip(a, b).

2. Step 2: Check each option

   let zipped = zip(a, b) uses correct syntax. Options B, C, and D use invalid syntax.

3. Final Answer:

   let zipped = zip(a, b) -> Option A

4. Quick Check:

   Function call uses parentheses = A [OK]

Hint: Use parentheses for function calls like zip(a, b) [OK]

Common Mistakes:

• Using square brackets instead of parentheses
• Trying to call zip as a method on array
• Omitting parentheses around arguments

3. What is the output of this Swift code?

let numbers = [1, 2, 3]
let letters = ["a", "b", "c", "d"]
for pair in zip(numbers, letters) {
    print(pair, terminator: " ")
}

medium

A. [(1, "a"), (2, "b"), (3, "c"), (nil, "d")]

B. (1, "a") (2, "b") (3, "c")

C. (1, "a") (2, "b") (3, "c") (4, "d")

D. (1, "a") (2, "b") (3, "c") (nil, nil)

Solution

1. Step 1: Understand zip behavior with different lengths

   Zip stops at the shortest sequence length, which is 3 here.

2. Step 2: Trace the loop output

   Pairs printed are (1, "a"), (2, "b"), (3, "c") only.

3. Final Answer:

   (1, "a") (2, "b") (3, "c") -> Option B

4. Quick Check:

   Zip stops at shortest = B [OK]

Hint: Zip stops at shortest sequence length, ignore extra elements [OK]

Common Mistakes:

• Expecting zip to include extra elements from longer sequence
• Thinking zip fills missing values with nil
• Confusing tuple output format

4. Identify the error in this Swift code using zip:

let nums = [1, 2, 3]
let chars = ["x", "y"]
for (num, char) in zip(nums, chars) {
    print(num + char)
}

medium

A. Missing parentheses in for loop

B. zip requires sequences of the same length

C. Cannot add Int and String directly

D. zip cannot be used with arrays

Solution

1. Step 1: Analyze the print statement

   The code tries to add an Int and a String with num + char, which is invalid.

2. Step 2: Check zip usage

   Zip works fine with arrays of different lengths; it stops at shortest.

3. Final Answer:

   Cannot add Int and String directly -> Option C

4. Quick Check:

   Type mismatch in addition = C [OK]

Hint: Check data types before adding values inside zip loop [OK]

Common Mistakes:

• Assuming zip needs equal length sequences
• Ignoring type mismatch errors
• Thinking zip syntax is wrong here

5. Given two arrays names = ["Anna", "Bob", "Cara"] and scores = [85, 92, 78], which Swift code correctly creates a dictionary pairing each name with its score using zip?

hard

A. let result = Dictionary(uniqueKeysWithValues: zip(names, scores))

B. let result = zip(names, scores).toDictionary()

C. let result = Dictionary(zip(names, scores))

D. let result = names.zip(scores).toDict()

Solution

1. Step 1: Understand how to convert zipped pairs to dictionary

   Use Dictionary(uniqueKeysWithValues:) initializer with zipped sequence.

2. Step 2: Check each option's validity

   let result = Dictionary(uniqueKeysWithValues: zip(names, scores)) uses correct syntax. Options B, C, and D use invalid or non-existent methods.

3. Final Answer:

   let result = Dictionary(uniqueKeysWithValues: zip(names, scores)) -> Option A

4. Quick Check:

   Use Dictionary initializer with zip = A [OK]

Hint: Use Dictionary(uniqueKeysWithValues:) with zip to create dictionary [OK]

Common Mistakes:

• Trying to call non-existent methods like toDictionary()
• Passing zip directly to Dictionary() without initializer
• Using zip as a method on array