object Anagrams {
/** A word is simply a `String`. */
type Word = String
/** A sentence is a `List` of words. */
type Sentence = List[Word]
/** `Occurrences` is a `List` of pairs of characters and positive integers saying
* how often the character appears.
* This list is sorted alphabetically w.r.t. to the character in each pair.
* All characters in the occurrence list are lowercase.
*
* Any list of pairs of lowercase characters and their frequency which is not sorted
* is **not** an occurrence list.
*
* Note: If the frequency of some character is zero, then that character should not be
* in the list.
*/
type Occurrences = List[(Char, Int)]
/** The dictionary is simply a sequence of words.
* It is predefined and obtained as a sequence using the utility method `loadDictionary`.
*/
val dictionary: List[Word] = loadDictionary
/** Converts the word into its character occurrence list.
*
* Note: the uppercase and lowercase version of the character are treated as the
* same character, and are represented as a lowercase character in the occurrence list.
*
* Note: you must use `groupBy` to implement this method!
*/
def wordOccurrences(w: Word): Occurrences = w.toLowerCase.groupBy(c => c).map(c => Occurrences((c._1, c._2.length)))
/** Converts a sentence into its character occurrence list. */
def sentenceOccurrences(s: Sentence): Occurrences = s.flatMap(w => wordOccurrences(w))
/** The `dictionaryByOccurrences` is a `Map` from different occurrences to a sequence of all
* the words that have that occurrence count.
* This map serves as an easy way to obtain all the anagrams of a word given its occurrence list.
*
* For example, the word "eat" has the following character occurrence list:
*
* `List(('a', 1), ('e', 1), ('t', 1))`
*
* Incidentally, so do the words "ate" and "tea".
*
* This means that the `dictionaryByOccurrences` map will contain an entry:
*
* List(('a', 1), ('e', 1), ('t', 1)) -> Seq("ate", "eat", "tea")
*
*/
lazy val dictionaryByOccurrences: Map[Occurrences, List[Word]] = dictionary.groupBy(wordOccurrences)
/** Returns all the anagrams of a given word. */
def wordAnagrams(word: Word): List[Word] = dictionaryByOccurrences.getOrElse(wordOccurrences(word), Nil)
/** Returns the list of all subsets of the occurrence list.
* This includes the occurrence itself, i.e. `List(('k', 1), ('o', 1))`
* is a subset of `List(('k', 1), ('o', 1))`.
* It also include the empty subset `List()`.
*
* Example: the subsets of the occurrence list `List(('a', 2), ('b', 2))` are:
*
* List(
* List(),
* List(('a', 1)),
* List(('a', 2)),
* List(('b', 1)),
* List(('a', 1), ('b', 1)),
* List(('a', 2), ('b', 1)),
* List(('b', 2)),
* List(('a', 1), ('b', 2)),
* List(('a', 2), ('b', 2))
* )
*
* Note that the order of the occurrence list subsets does not matter -- the subsets
* in the example above could have been displayed in some other order.
*/
def combinations(occurrences: Occurrences): List[Occurrences] =
(occurrences foldRight List[Occurrences](Nil)) { case ((ch,tm), acc) => {
acc ++ ( for { comb <- acc; n <- 1 to tm } yield (ch, n) :: comb )
} }
/** Subtracts occurrence list `y` from occurrence list `x`.
*
* The precondition is that the occurrence list `y` is a subset of
* the occurrence list `x` -- any character appearing in `y` must
* appear in `x`, and its frequency in `y` must be smaller or equal
* than its frequency in `x`.
*
* Note: the resulting value is an occurrence - meaning it is sorted
* and has no zero-entries.
*/
def subtract(x: Occurrences, y: Occurrences): Occurrences = {
(y foldLeft SortedMap[Char,Int]() ++ x){ case (map, (ch, tm)) => {
val newTm = map(ch) - tm
if (newTm != 0) map updated (ch, newTm)
else map - ch
} }.toList
}
/** Returns a list of all anagram sentences of the given sentence.
*
* An anagram of a sentence is formed by taking the occurrences of all the characters of
* all the words in the sentence, and producing all possible combinations of words with those characters,
* such that the words have to be from the dictionary.
*
* The number of words in the sentence and its anagrams does not have to correspond.
* For example, the sentence `List("I", "love", "you")` is an anagram of the sentence `List("You", "olive")`.
*
* Also, two sentences with the same words but in a different order are considered two different anagrams.
* For example, sentences `List("You", "olive")` and `List("olive", "you")` are different anagrams of
* `List("I", "love", "you")`.
*
* Here is a full example of a sentence `List("Yes", "man")` and its anagrams for our dictionary:
*
* List(
* List(en, as, my),
* List(en, my, as),
* List(man, yes),
* List(men, say),
* List(as, en, my),
* List(as, my, en),
* List(sane, my),
* List(Sean, my),
* List(my, en, as),
* List(my, as, en),
* List(my, sane),
* List(my, Sean),
* List(say, men),
* List(yes, man)
* )
*
* The different sentences do not have to be output in the order shown above - any order is fine as long as
* all the anagrams are there. Every returned word has to exist in the dictionary.
*
* Note: in case that the words of the sentence are in the dictionary, then the sentence is the anagram of itself,
* so it has to be returned in this list.
*
* Note: There is only one anagram of an empty sentence.
*/
def sentenceAnagrams(sentence: Sentence): List[Sentence] = {
def iter(occurrences: Occurrences): List[Sentence] = {
if (occurrences.isEmpty) List(Nil)
else for {
combination <- combinations( occurrences )
word <- dictionaryByOccurrences getOrElse (combination, Nil)
sentence <- iter( subtract(occurrences,wordOccurrences(word)) )
if !combination.isEmpty
} yield word :: sentence
}
iter( sentenceOccurrences(sentence) )
}
}
