Generating Phone Words in Swift

PhoneWords.swift

/*: 
## Phone Words

Generate a collection of words that can be represented by a given phone number. If a phone number contains the digits `1` or `0` then split up the phone number and find the words for each of the substrings as long as each substring has more than one digit. Non-keypad characters can be ignored. Optionally, filter out words so that only dictionary words are present in the result.

    ╔═════╦═════╦═════╗
    ║  1  ║  2  ║  3  ║
    ║     ║ abc ║ def ║
    ╠═════╬═════╬═════╣
    ║  4  ║  5  ║  6  ║
    ║ ghi ║ jkl ║ mno ║
    ╠═════╬═════╬═════╣
    ║  7  ║  8  ║  9  ║
    ║ pqrs║ tuv ║wxyz ║
    ╠═════╬═════╬═════╣
    ║  *  ║  0  ║  #  ║
    ║     ║     ║     ║
    ╚═════╩═════╩═════╝

Example

    `1-8000-356-9377` should return "FLOWERS"
*/

/*
The built-in words file "/usr/share/dict/words" can also be used. But it seems to be missing plurals.
The "words.txt" file used below can be found at [dwyl/english-words](https://github.com/dwyl/english-words).

*If you are pasting this into a playground then it's best to put this part in the "Sources" directory so it doesn't get evaluated every time*
*/
let words = Set(
    try! String(contentsOfURL: NSBundle.mainBundle().URLForResource("words", withExtension: "txt")!)
        .componentsSeparatedByCharactersInSet(
            NSCharacterSet.newlineCharacterSet()
    )
)

public func isWord(s: String) -> Bool {
    return words.contains(s)
}

/*:
This solution involves using a map of `Character` to `String` arrays. `String` arrays are used instead of just strings since we have to split each string into characters and convert back to `String` for concatenation anyway.
*/
let phoneKeyMap: [Character: [String]] = [
    "2": ["a","b","c"],
    "3": ["d","e","f"],
    "4": ["g","h","i"],
    "5": ["j","k","l"],
    "6": ["m","n","o"],
    "7": ["p","q","r","s"],
    "8": ["t","u","v"],
    "9": ["w","x","y","z"]
]

/*:
Generate phone words recursively

1. handle empty string
2. handle non keypad characters
3. termination condition
4. recursively build words
5. combine words with all characters for cur key
*/
func phoneWords(s: String) -> [String] {
    guard let n = s.characters.first else { return [] } // 1
    guard let px = phoneKeyMap[n] else { return [] } // 2
    if s.characters.count == 1 { return px } // 3
    let sx = phoneWords(String(s.characters.dropFirst())) // 4
    return px.flatMap { p in sx.map { s in p + s } } // 5
}

/*:
Some helper functions to make things more readable
*/

func isPhoneKeyMappable(c: Character) -> Bool {
    return ("2"..."9").contains(String(c))
}

func hasMoreThanOne<T: CollectionType>(c: T) -> Bool {
    return c.count > 1
}

func filterWith<T: CollectionType>(pred: T.Generator.Element -> Bool)(c: T) -> [T.Generator.Element] {
    return c.filter(pred)
}

/*:
Generate phone words for the phone number `1-800-FLOWERS`

1. split the phone number at "1" and "0"
2. filter out digits that are not 2-9
3. filter out single-digit words
4. convert each collection of `Character` into a `String`
5. convert each string into a collection of words based on keypad
6. filter out non-engligh words based on a word dictionary
*/
let mnemonics = "1-800-356-9377".characters
    .split { ("0"..."1").contains(String($0)) } // 1
    .map(filterWith(isPhoneKeyMappable)) // 2
    .filter(hasMoreThanOne) // 3
    .map(String.init) // 4
    .map(phoneWords) // 5
    .map(filterWith(isWord)) // 6

print(mnemonics) // [["flowers"]]

/*: 
Do the reverse; convert a mnemonic phone word to its phone number.
*/
func phoneNumber(s: String) -> String {
    return s.lowercaseString.characters.map {
        switch $0 {
        case "a"..."c": return "2"
        case "d"..."f": return "3"
        case "g"..."i": return "4"
        case "j"..."l": return "5"
        case "m"..."o": return "6"
        case "p"..."s": return "7"
        case "t"..."v": return "8"
        case "w"..."z": return "9"
        default: return String($0)
        }
    }.reduce("", combine: +)
}

phoneNumber("1-800-FLOWERS") // "1-800-3569377"

// I do not endorse 1-800-FLOWERS