Interview practice using Swift

WordCount.swift

import Foundation

// Write a function in swift that takes a sentence as input and returns the count of each word in the sentence. Ignore punctuation and consider words as case-insensitive. For example, given the input "Hello, world! hello", the function should return ["hello": 2, "world": 1]

func wordCount(str: String) -> [String: Int] {
    // if empty input string, return empty dict
    guard !str.isEmpty else {
        return [:]
    }
    
    // what is the minimum word count? can we accept "a" or "I"? yes
    
    // what is character limit of input string?
    
    // what would be a legit word? "asdf asdf asdf liuhj ❌🔥" [a-zA-Z]
    
    // 1. remove the punctuation
    // 2. lowercase all letters
    
    var string = str
        .components(separatedBy: .punctuationCharacters)
        .joined(separator: "")
        .components(separatedBy: .symbols)
        .joined(separator: "")
        .lowercased()
    
    // 3. split the words by whitespace -> array
    var words = string.components(separatedBy: .whitespaces)
    var dict: [String: Int] = [:]
    
    // 4. go through array of words
    for word in words {
        guard !word.isEmpty else {
            continue
        }
        
        //     - if the word exists in the dict, then we increment the value (+1)
        if let wordCount = dict[word] {
            dict[word] = wordCount + 1
        //     - if we see a new word, then we insert that word into dict with value 1
        } else {
            dict[word] = 1
        }
    }
    return dict
}

var greeting = "Hello, playground"

let test1 = wordCount(str: "Hello, world! hello")
print(test1) // ["hello": 2, "world": 1]

let test2 = wordCount(str: "")
print(test2) // [:]

let test3 = wordCount(str: "!@#*&$% !@#&$%^ ())*%-=")
print(test3) // [:]