Dictionary creation and access in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift dictionary access?

Consider the following Swift code. What will be printed?

Swift

let fruits = ["apple": 3, "banana": 5, "orange": 2]
print(fruits["banana"] ?? 0)

A 0

B Optional(5)

C banana

D 5

💡 Hint

Remember that using ?? provides a default value if the key is missing.

The dictionary fruits has a key "banana" with value 5. Accessing fruits["banana"] returns an Optional Int. Using ?? 0 unwraps it or returns 0 if nil. So the output is 5.

❓ Predict Output

intermediate

What does this dictionary creation code produce?

What is the content of the dictionary 'numbers' after this code runs?

Swift

let numbers = Dictionary(uniqueKeysWithValues: [(1, "one"), (2, "two"), (3, "three")])
print(numbers)

A Error: Cannot create dictionary this way

B [1: "one", 2: "two", 3: "three"]

C [(1, "one"), (2, "two"), (3, "three")]

D ["one": 1, "two": 2, "three": 3]

💡 Hint

Dictionary(uniqueKeysWithValues:) creates a dictionary from a sequence of key-value pairs.

The code creates a dictionary with keys 1, 2, 3 and values "one", "two", "three" respectively. The print shows the dictionary in key:value pairs.

❓ Predict Output

advanced

What error does this Swift dictionary code raise?

What error will occur when running this code?

Swift

var dict = ["a": 1, "b": 2]
let value = dict["c"]!
print(value)

A Runtime error: Unexpectedly found nil while unwrapping an Optional value

B Compile-time error: Key not found in dictionary

C Prints 0

D Prints nil

💡 Hint

Using ! to unwrap an Optional that is nil causes a runtime crash.

Accessing dict["c"] returns nil because "c" is not a key. Using ! to force unwrap nil causes a runtime crash with the given error.

🧠 Conceptual

advanced

How many items are in the dictionary after this code?

After running this Swift code, how many key-value pairs does the dictionary 'items' contain?

Swift

var items = ["x": 10, "y": 20]
items["z"] = 30
items["x"] = nil

A 1

B 3

C 2

D 0

💡 Hint

Setting a dictionary value to nil removes the key.

Initially, items has 2 keys: "x" and "y". Adding "z" makes 3 keys. Setting "x" to nil removes it, leaving 2 keys: "y" and "z".

🔧 Debug

expert

What is the output of this Swift dictionary comprehension?

What will be printed by this Swift code?

Swift

let squares = Dictionary(uniqueKeysWithValues: (1...3).map { ($0, $0 * $0) })
print(squares[2] ?? -1)

A 4

B 2

C -1

D Optional(4)

💡 Hint

The dictionary maps numbers to their squares. Accessing key 2 returns 4.

The dictionary squares has keys 1, 2, 3 with values 1, 4, 9. Accessing squares[2] returns Optional(4). Using ?? -1 unwraps it to 4.

Practice

1. What is a dictionary in Swift used for?

easy

A. To store data in key-value pairs for quick lookup

B. To store a list of values without keys

C. To perform mathematical calculations

D. To create user interface elements

Solution

1. Step 1: Understand dictionary purpose

   Dictionaries store data as pairs of keys and values, allowing fast access by key.

2. Step 2: Compare options

   Only To store data in key-value pairs for quick lookup correctly describes this key-value storage purpose.

3. Final Answer:

   To store data in key-value pairs for quick lookup -> Option A

4. Quick Check:

   Dictionaries = key-value storage [OK]

Hint: Remember: dictionaries pair keys with values for fast access [OK]

Common Mistakes:

• Confusing dictionaries with arrays (which have no keys)
• Thinking dictionaries perform calculations
• Assuming dictionaries create UI elements

2. Which of the following is the correct way to create a dictionary in Swift?

easy

A. let dict = ("apple", 1), ("banana", 2)

B. let dict = ["apple": 1, "banana": 2]

C. let dict = {"apple": 1, "banana": 2}

D. let dict = ["apple", 1, "banana", 2]

Solution

1. Step 1: Identify Swift dictionary syntax

   Swift dictionaries use square brackets with key-value pairs separated by colons.

2. Step 2: Check each option

   let dict = ["apple": 1, "banana": 2] uses correct syntax: ["key": value, ...]. Others use tuples, braces, or arrays incorrectly.

3. Final Answer:

   let dict = ["apple": 1, "banana": 2] -> Option B

4. Quick Check:

   Dictionary syntax = [key: value] [OK]

Hint: Use square brackets with colons for key-value pairs [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Using curly braces which define closures
• Listing keys and values without colons

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

let fruits = ["apple": 3, "banana": 5]
print(fruits["banana"] ?? 0)

medium

A. 5

B. nil

C. banana

D. 0

Solution

1. Step 1: Access dictionary value by key

   fruits["banana"] looks up the value for key "banana", which is 5.

2. Step 2: Use nil-coalescing operator

   The ?? 0 means if the key is missing (nil), print 0; here key exists, so print 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   fruits["banana"] = 5 [OK]

Hint: Use ?? to provide default if key missing [OK]

Common Mistakes:

• Expecting nil instead of default value
• Printing the key instead of the value
• Confusing optional with direct value

4. Identify the error in this Swift code snippet:

var scores = ["Alice": 90, "Bob": 85]
scores["Charlie"] = 95
print(scores["Charlie"])

medium

A. Cannot add new key-value pairs to a dictionary

B. Dictionary keys must be integers

C. Syntax error in dictionary declaration

D. Missing unwrap of optional when printing

Solution

1. Step 1: Check dictionary modification

   Adding a new key-value pair to a var dictionary is allowed, so no error there.

2. Step 2: Understand optional access

   Accessing scores["Charlie"] returns an optional Int?; printing it directly shows Optional(95), not 95.

3. Step 3: Identify missing unwrap

   To print the value without Optional(...), unwrap it safely or use ?? to provide default.

4. Final Answer:

   Missing unwrap of optional when printing -> Option D

5. Quick Check:

   Dictionary access returns optional, unwrap needed [OK]

Hint: Remember dictionary access returns optional, unwrap before use [OK]

Common Mistakes:

• Thinking keys must be integers
• Assuming dictionary is immutable by default
• Ignoring optional nature of dictionary values

5. Given a dictionary let items = ["pen": 3, "notebook": 0, "eraser": 2], which Swift code correctly creates a new dictionary containing only items with quantity greater than zero?

hard

A. let filtered = items.compactMap { $0.value > 0 }

B. let filtered = items.filter { $0.key > 0 }

C. let filtered = items.filter { $0.value > 0 }

D. let filtered = items.map { $0.value > 0 }

Solution

1. Step 1: Understand filtering dictionary by value

   We want to keep pairs where value > 0, so filter with condition on $0.value.

2. Step 2: Analyze each option

   let filtered = items.filter { $0.value > 0 } uses filter with correct condition. let filtered = items.filter { $0.key > 0 } incorrectly compares key (a String) to 0. Options A and C use map/compactMap which transform values, not filter pairs.

3. Final Answer:

   let filtered = items.filter { $0.value > 0 } -> Option C

4. Quick Check:

   Filter dictionary by value > 0 = let filtered = items.filter { $0.value > 0 } [OK]

Hint: Use filter with $0.value condition to keep matching pairs [OK]

Common Mistakes:

• Filtering by key instead of value
• Using map instead of filter for selection
• Using compactMap which changes type, not filters