Zip for combining sequences in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to combine two arrays using zip and print the pairs.

Swift

let numbers = [1, 2, 3]
let words = ["one", "two", "three"]
for pair in [1](numbers, words) {
    print(pair)
}

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

Amerge

Bcombine

Czip

Djoin

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'combine' or 'merge' which are not Swift functions for this purpose.

The zip function combines two sequences into pairs.

2fill in blank

medium

Complete the code to print the first elements of each pair from zipped arrays.

Swift

let a = [10, 20, 30]
let b = ["A", "B", "C"]
for ([1], _) in zip(a, b) {
    print([1])
}

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

Ax

Bfirst

Celement

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'first' which is a property, not a variable name here.

The variable 'x' is used to capture the first element in the tuple.

3fill in blank

hard

Fix the error in the code to correctly zip and print pairs.

Swift

let nums = [1, 2, 3]
let strs = ["one", "two", "three"]
for pair in zip(nums, [1]) {
    print(pair)
}

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

Astrs

Bstrings

Cstr

Dstrs()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'strings' or 'strs()' which are undefined or incorrect syntax.

The variable 'strs' holds the second array to zip with 'nums'.

4fill in blank

hard

Fill both blanks to create a dictionary from zipped arrays where keys are numbers and values are words.

Swift

let keys = [1, 2, 3]
let values = ["one", "two", "three"]
let dict = Dictionary(uniqueKeysWithValues: [1](keys, [2]))

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

Azip

Bvalues

Ckeys

Denumerated

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'enumerated' which is not for pairing two sequences.

Use zip to pair keys and values arrays to create the dictionary.

5fill in blank

hard

Fill all three blanks to filter zipped pairs where the number is greater than 1 and print the word.

Swift

let nums = [1, 2, 3]
let words = ["one", "two", "three"]
for ([1], [2]) in zip(nums, words) where [1] > 1 {
    print([2])
}

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

Anum

Bword

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names or wrong variables in the condition.

Use 'num' and 'word' as variable names; filter where num > 1 and print word.

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