Dictionary methods and default values in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the value for key "apple" from the dictionary.

Swift

let fruits = ["apple": 3, "banana": 5]
let count = fruits[[1]]

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

A"orange"

B"apple"

C"banana"

D"grape"

Attempts:

3 left

💡 Hint

Common Mistakes

Using a key that does not exist in the dictionary.

Forgetting to put the key inside quotes.

The key "apple" is used to access its value in the dictionary.

2fill in blank

medium

Complete the code to provide a default value of 0 when the key is missing.

Swift

let fruits = ["apple": 3, "banana": 5]
let count = fruits["orange", default: [1]]

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

Anil

B5

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using nil as a default value which is not allowed here.

Using a default value that does not match the dictionary's value type.

Using default: 0 returns 0 if the key is not found.

3fill in blank

hard

Fix the error in updating the dictionary value safely.

Swift

var fruits = ["apple": 3, "banana": 5]
fruits["apple"] = fruits["apple"]! [1] 2

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Forgetting to unwrap the optional value before adding.

We add 2 to the existing value for key "apple".

4fill in blank

hard

Fill both blanks to create a dictionary with squares of numbers greater than 3.

Swift

let squares = [[1]: [2] for num in 1...5 if num > 3]

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

Anum

Bnum * num

Cnum + num

Dnum - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or subtraction instead of multiplication for squares.

Using the wrong variable as key or value.

The dictionary keys are numbers and values are their squares for numbers greater than 3.

5fill in blank

hard

Fill all three blanks to filter dictionary items with values greater than 4 and map keys to uppercase.

Swift

let fruits = ["apple": 3, "banana": 5, "cherry": 7]
let filtered = [[1]: [2] for (key, value) in fruits where value [3] 4]

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

Akey.uppercased()

Bvalue

C>

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Not converting keys to uppercase.

Using wrong comparison operator or value.

Keys are converted to uppercase, values are kept, and only items with values greater than 4 are included.

Practice

1. What does the updateValue(_:forKey:) method do in a Swift dictionary?

easy

A. It updates the value for a given key and returns the old value if it existed.

B. It removes the key and its value from the dictionary.

C. It returns the value for a key or a default if the key is missing.

D. It creates a new dictionary with updated keys.

Solution

1. Step 1: Understand updateValue method purpose

   The updateValue(_:forKey:) method changes the value for a specified key in the dictionary.

2. Step 2: Check return behavior

   This method returns the old value if the key existed, otherwise nil.

3. Final Answer:

   It updates the value for a given key and returns the old value if it existed. -> Option A

4. Quick Check:

   updateValue returns old value or nil [OK]

Hint: Remember updateValue returns old value if key exists [OK]

Common Mistakes:

• Confusing updateValue with removeValue
• Thinking it returns the new value
• Assuming it creates a new dictionary

2. Which of the following is the correct syntax to get a value from a dictionary with a default in Swift?

easy

A. let value = dict[key, default: 0]

B. let value = dict.get(key, default: 0)

C. let value = dict[key] ?? 0

D. let value = dict.value(forKey: key, default: 0)

Solution

1. Step 1: Recall Swift dictionary default value syntax

   Swift dictionaries allow accessing values with a default using dict[key, default: defaultValue].

2. Step 2: Compare options

   let value = dict[key, default: 0] matches the correct syntax. let value = dict[key] ?? 0 uses nil-coalescing but is not the dictionary default syntax.

3. Final Answer:

   let value = dict[key, default: 0] -> Option A

4. Quick Check:

   Use dict[key, default:] for default values [OK]

Hint: Use square brackets with default: for safe dictionary access [OK]

Common Mistakes:

• Using get() method which doesn't exist
• Confusing nil-coalescing with dictionary default syntax
• Using value(forKey:) which is for NSObject

3. What is the output of this Swift code?

var scores = ["Alice": 10, "Bob": 8]
let oldValue = scores.updateValue(12, forKey: "Alice")
let missingValue = scores.updateValue(5, forKey: "Charlie")
print(oldValue, missingValue, scores)

medium

A. nil Optional(5) ["Alice": 12, "Bob": 8, "Charlie": 5]

B. 10 5 ["Alice": 12, "Bob": 8, "Charlie": 5]

C. Optional(10) nil ["Alice": 12, "Bob": 8, "Charlie": 5]

D. Optional(12) Optional(5) ["Alice": 10, "Bob": 8, "Charlie": 5]

Solution

1. Step 1: Understand updateValue return values

   Updating "Alice" from 10 to 12 returns old value Optional(10). Adding "Charlie" returns nil because key was missing.

2. Step 2: Check dictionary after updates

   Dictionary now has "Alice":12, "Bob":8, and "Charlie":5.

3. Final Answer:

   Optional(10) nil ["Alice": 12, "Bob": 8, "Charlie": 5] -> Option C

4. Quick Check:

   updateValue returns old value or nil [OK]

Hint: updateValue returns old value wrapped in Optional or nil if new key [OK]

Common Mistakes:

• Expecting updateValue to return new value
• Forgetting Optional wrapping
• Not updating dictionary correctly

4. Identify the error in this Swift code snippet:

var dict = ["a": 1, "b": 2]
let removed = dict.removeValue(forKey: "c")
print(removed!)

medium

A. Dictionary keys must be integers, not strings.

B. Force unwrapping removed causes runtime crash if key missing.

C. removeValue(forKey:) does not exist for dictionaries.

D. print statement syntax is incorrect.

Solution

1. Step 1: Understand removeValue return type

   removeValue(forKey:) returns an Optional value. If key is missing, it returns nil.

2. Step 2: Analyze force unwrapping

   Force unwrapping removed! when key "c" is missing causes a runtime crash.

3. Final Answer:

   Force unwrapping removed causes runtime crash if key missing. -> Option B

4. Quick Check:

   Force unwrap nil causes crash [OK]

Hint: Avoid force unwrapping optional from removeValue without check [OK]

Common Mistakes:

• Assuming removeValue always returns a value
• Ignoring optional unwrapping
• Thinking removeValue removes all keys

5. You have a dictionary var inventory = ["apple": 3, "banana": 5]. Write code to safely add 2 to the count of "orange", even if it doesn't exist yet, using dictionary default values. What is the correct way?

hard

A. inventory["orange"] = inventory["orange"]! + 2

B. inventory["orange"] = inventory["orange"] ?? 0 + 2

C. inventory.updateValue(inventory["orange"]! + 2, forKey: "orange")

D. inventory["orange"] = inventory["orange", default: 0] + 2

Solution

1. Step 1: Understand safe addition with default

   Using inventory["orange", default: 0] returns 0 if "orange" key is missing.

2. Step 2: Increment value safely

   Adding 2 to this default value and assigning back to inventory["orange"] updates or creates the key safely.

3. Final Answer:

   inventory["orange"] = inventory["orange", default: 0] + 2 -> Option D

4. Quick Check:

   dict[key] = dict[key, default: 0] + value for safe increments [OK]

Hint: Use dict[key] = dict[key, default: 0] + amount to add safely [OK]

Common Mistakes:

• Force unwrapping missing keys causing crash
• Using += on read-only default subscript
• Using updateValue without checking for nil