Why collection algorithms matter in Swift - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the first element in the array.

Swift

let numbers = [1, 2, 3, 4, 5]
let firstNumber = numbers.[1]

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

Alast()

Bfirst()

Ccount

Dappend()

Attempts:

3 left

💡 Hint

Common Mistakes

Using last() returns the last element, not the first.

Using count returns the number of elements, not an element.

append() adds an element, it does not retrieve one.

The first() method returns the first element of the array.

2fill in blank

medium

Complete the code to filter numbers greater than 3.

Swift

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

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

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' filters numbers less than 3, not greater.

Using '==' filters only numbers equal to 3.

Using '<=' includes numbers less than or equal to 3.

The filter method keeps elements where the condition is true. Here, we want numbers greater than 3.

3fill in blank

hard

Fix the error in the code to sum all numbers in the array.

Swift

let numbers = [1, 2, 3, 4, 5]
let total = numbers.reduce(0, [1])

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

A{ $1 * $0 }

B{ $0 / $1 }

C{ $0 - $1 }

D{ $0 + $1 }

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication changes the result.

Division can cause runtime errors if dividing by zero.

The reduce method combines all elements using the closure. To sum, add the accumulator and current value.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths for words longer than 3 characters.

Swift

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

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

A>

Bcount

Clength

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' filters shorter words, not longer.

Using 'length' is not a valid property in Swift strings.

Filter words with count greater than 3, then map each word to a tuple of (word, word.count).

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values greater than 2.

Swift

let data = ["one": 1, "two": 2, "three": 3]
let result = data.filter { $0.value [1] 2 }.mapKeys { $0.key.[2]() }.mapValues { $0.[3] }

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

A>

Buppercased

Cdescription

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' filters values less than 2.

Using 'lowercased' converts keys to lowercase, not uppercase.

Using 'count' on values is invalid since values are Int.

Filter values greater than 2, convert keys to uppercase, and convert values to strings using description.

Practice

1. Why are collection algorithms important when working with data in Swift?

easy

A. They help find, sort, and filter data easily and efficiently.

B. They make the app run slower by adding extra steps.

C. They only work with numbers, not text or objects.

D. They require complex code that is hard to read.

Solution

1. Step 1: Understand the purpose of collection algorithms

   Collection algorithms are designed to help manage data by finding, sorting, filtering, and checking items quickly.

2. Step 2: Evaluate the options based on this purpose

   They help find, sort, and filter data easily and efficiently. correctly states their role in making data handling easier and efficient. Other options describe incorrect or negative effects.

3. Final Answer:

   They help find, sort, and filter data easily and efficiently. -> Option A

4. Quick Check:

   Collection algorithms improve data handling = B [OK]

Hint: Remember: algorithms simplify data tasks fast [OK]

Common Mistakes:

• Thinking algorithms slow down the app
• Believing algorithms only work with numbers
• Assuming algorithms make code complex

2. Which of the following is the correct way to use the filter algorithm on an array numbers to get only even numbers in Swift?

easy

A. let evens = numbers.filter { $0 / 2 == 0 }

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

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

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

Solution

1. Step 1: Understand the filter syntax in Swift

   The filter method takes a closure that returns true for elements to keep. The condition for even numbers is number % 2 == 0.

2. Step 2: Check each option for correct syntax and logic

   let evens = numbers.filter { $0 % 2 == 0 } uses the correct modulo operator and comparison. let evens = numbers.filter { $0 / 2 == 0 } uses division incorrectly. let evens = numbers.filter { $0 % 2 = 0 } uses assignment (=) instead of comparison (==). let evens = numbers.filter { $0 == 2 } filters only numbers equal to 2, not all evens.

3. Final Answer:

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

4. Quick Check:

   Correct filter syntax for evens = A [OK]

Hint: Use % 2 == 0 to find even numbers in filter [OK]

Common Mistakes:

• Using / instead of % for even check
• Using = instead of == in condition
• Filtering only one number instead of all evens

3. What will be the output of this Swift code?

let names = ["Anna", "Bob", "Cathy", "David"]
let result = names.contains { $0.hasPrefix("C") }
print(result)

medium

A. nil

B. false

C. SyntaxError

D. true

Solution

1. Step 1: Understand the contains method with a closure

   The contains method checks if any element satisfies the condition. Here, it checks if any name starts with "C".

2. Step 2: Check the array elements for the condition

   "Cathy" starts with "C", so the condition is true, making result true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   names contains prefix "C" = true [OK]

Hint: contains with closure returns true if any match found [OK]

Common Mistakes:

• Thinking contains checks all elements and returns false if any fail
• Confusing hasPrefix with contains
• Expecting syntax error due to closure

4. Identify the error in this Swift code that tries to sort an array of integers:

let numbers = [3, 1, 4, 2]
let sorted = numbers.sorted(by: { a, b in a > b })
print(sorted)

medium

A. The closure should use '<' instead of '>' to sort ascending.

B. The sorted method cannot take a closure.

C. The array must be mutable to sort.

D. The closure parameters must be named 'x' and 'y'.

Solution

1. Step 1: Understand the sorted(by:) closure logic

   The closure defines the sorting order. Using 'a > b' sorts descending, not ascending.

2. Step 2: Identify the intended sort order

   Since the question implies sorting ascending, the closure should use 'a < b'. Other options are incorrect because sorted can take closures, arrays can be immutable for sorted, and parameter names are flexible.

3. Final Answer:

   The closure should use '<' instead of '>' to sort ascending. -> Option A

4. Quick Check:

   Use '<' for ascending sort closure [OK]

Hint: Use '<' in closure for ascending sort [OK]

Common Mistakes:

• Thinking sorted needs mutable array
• Believing closure parameter names are fixed
• Assuming sorted can't take closures

5. You have a list of user scores: let scores = [10, 0, 5, 0, 8]. You want to create a dictionary where keys are the scores and values are the count of how many times each score appears, but you want to ignore zero scores. Which Swift code correctly uses collection algorithms to do this?

hard

A. let counts = scores.filter { $0 == 0 }.count

B. let counts = Dictionary(uniqueKeysWithValues: scores.map { ($0, 1) })

C. let counts = Dictionary(grouping: scores.filter { $0 != 0 }, by: { $0 }).mapValues { $0.count }

D. let counts = scores.reduce(into: [:]) { $0[$1] = 1 }

Solution

1. Step 1: Filter out zero scores

   The code filters scores to exclude zeros using filter { $0 != 0 }.

2. Step 2: Group remaining scores and count occurrences

   Using Dictionary(grouping:by:) groups scores by value, then mapValues { $0.count } counts how many times each score appears.

3. Final Answer:

   let counts = Dictionary(grouping: scores.filter { $0 != 0 }, by: { $0 }).mapValues { $0.count } -> Option C

4. Quick Check:

   Filter zeros, group by score, count occurrences = D [OK]

Hint: Filter zeros first, then group and count [OK]

Common Mistakes:

• Using uniqueKeysWithValues which overwrites duplicates
• Not filtering zeros before counting
• Using reduce incorrectly without counting