Given an array words = ["swift", "code", "test"], how can you create a dictionary where keys are indices and values are the uppercase words using enumerated()?
hard📝 Application Q8 of 15
Swift - Collections
Given an array words = ["swift", "code", "test"], how can you create a dictionary where keys are indices and values are the uppercase words using enumerated()?
Avar dict = [Int: String]()
for (i, word) in words.enumerated() {
dict[i] = word.uppercased()
}
Bvar dict = [String: Int]()
for (i, word) in words.enumerated() {
dict[word] = i
}
Cvar dict = [Int: String]()
for word in words {
dict[word.count] = word
}
Dvar dict = [Int: String]()
for (word, i) in words.enumerated() {
dict[i] = word.lowercased()
}
Step-by-Step Solution
Solution:
Step 1: Understand desired dictionary structure
Keys are indices (Int), values are uppercase words (String).
Step 2: Check each option's logic
var dict = [Int: String]()
for (i, word) in words.enumerated() {
dict[i] = word.uppercased()
} correctly uses enumerated() to get index and word, then stores uppercase word with index key. var dict = [String: Int]()
for (i, word) in words.enumerated() {
dict[word] = i
} reverses key-value types. var dict = [Int: String]()
for word in words {
dict[word.count] = word
} uses word count as key incorrectly. var dict = [Int: String]()
for (word, i) in words.enumerated() {
dict[i] = word.lowercased()
} swaps tuple order and uses lowercased().
Final Answer:
var dict = [Int: String]()
for (i, word) in words.enumerated() {
dict[i] = word.uppercased()
} -> Option A
Quick Check:
Use enumerated() to map index to uppercase word [OK]
Quick Trick:Use enumerated() to get index and transform values [OK]
Common Mistakes:
Swapping key and value types
Using wrong tuple order
Not transforming word case
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