When you're sitting in a coding interview, the pressure often pushes you toward the first solution that comes to mind. That's usually the "naive" approach—the one that solves the problem but does so inefficiently. In a real interview, getting a working solution is great, but the magic happens when you can look at your own code and say, "This works, but it's slow; here is how I'd actually build it for production."

Let's walk through a classic: Grouping Anagrams. Given an array of strings, we need to group the ones that are anagrams of each other (meaning they use the exact same letters in a different order) into separate lists.

The "Just Make it Work" Approach

If I'm nervous and just trying to get something on the screen, I might start with a brute-force method. I'd probably iterate through the list, pick a word, and then scan the rest of the list to see if any other words match it. To check if two words are anagrams, I'd sort both strings and see if they're equal. It looks something like this:

func groupAnagramsNaive(_ strings: [String]) -> [[String]] {
    var result = [[String]]()
    var visited = Set<Int>()
    
    for i in 0..

Now, if you write this, the interviewer won't tell you it's "wrong" because it produces the correct output. But they are waiting for you to realize that this is an $O(N^2 \cdot K \log K)$ disaster, where $N$ is the number of strings and $K$ is the average length of a string. We are sorting strings inside a nested loop. If the input list has 10,000 words, this code is going to hang. I've seen candidates get stuck here because they're too afraid to admit their first attempt is inefficient.

Trading Space for Speed

The way to fix this is to stop comparing every word to every other word. Instead, we can use a Dictionary to act as a lookup table. The trick is to realize that every anagram, once sorted, results in the exact same string. That sorted string becomes our "canonical key."

By using the sorted string as a key and an array of original strings as the value, we only have to iterate through the list once. We've traded a bit of memory (the dictionary) for a massive gain in speed. We've moved from quadratic time to linear time relative to the number of strings.

func groupAnagramsOptimized(_ strings: [String]) -> [[String]] {
    var groups = [String: [String]]()
    
    for word in strings {
        let key = String(word.sorted())
        groups[key, default: []].append(word)
    }
    
    return Array(groups.values)
}

Notice the use of groups[key, default: []]. In Swift, this is much cleaner than checking if the key exists and then initializing an empty array. It's a small detail, but it shows the interviewer you actually know the language idioms.

The "Senior Engineer" Swift Shortcut

If you really want to impress, you can point out that Swift has a built-in initializer for dictionaries specifically for this purpose: Dictionary(grouping:by:). This is essentially a functional way of doing exactly what we did above. It tells the compiler, "Take this collection and group the elements based on the result of this closure."

func groupAnagramsSwiftWay(_ strings: [String]) -> [[String]] {
    let grouped = Dictionary(grouping: strings) { String($0.sorted()) }
    return Array(grouped.values)
}

I usually suggest presenting the Dictionary loop first, then mentioning the grouping initializer as a "Swift-specific optimization." It proves you understand the underlying data structure (the hash map) while also demonstrating that you're up-to-date with the Standard Library. It turns the conversation from "Can this person code?" to "How can this person help us write more concise Swift?"

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📋 Practical Task

Implement a High-Frequency Word Finder

To practice the pattern of using a dictionary to group and count data—similar to the anagram problem—write a function called findFrequentWords(in text: String, threshold: Int).

Your function should:

1. Take a long string of text and a minimum frequency count (threshold).
2. Clean the text by removing punctuation and converting it to lowercase.
3. Identify all words that appear more than the threshold number of times.
4. Return these words as an array of strings.

Constraint: Avoid using nested loops to count the words. Use a Dictionary to keep track of the counts in a single pass through the words array.