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.