package forcomp

import common._

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 occurence 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.
   */
  def wordOccurrences(w: Word): Occurrences = {
    val map = w.toLowerCase.toList.groupBy((c: Char) => c).map((t: (Char, List[Char])) => (t._1, t._2.length))
    map.toList.sortWith((f: (Char, Int), s: (Char, Int)) => f._1 < s._1)
  }

  /** Converts a sentence into its character occurrence list. */
  def sentenceOccurrences(s: Sentence): Occurrences = wordOccurrences(s.toList.flatten.mkString)

  /** 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((w: Word) => wordOccurrences(w)).withDefaultValue(List())
  }

  /** Returns all the anagrams of a given word. */
  def wordAnagrams(word: Word): List[Word] = dictionaryByOccurrences(wordOccurrences(word))

  /** 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] = {
    val nonSort = cartesianProd(occurrences.map((t: (Char, Int)) => combinate(t._1, t._2)))
    nonSort.map((x: List[(Char, Int)]) => x.sortWith((t1: (Char, Int), t2: (Char, Int)) => t1._1 < t2._2))
  }
  def combinate(char: Char, occ: Int): List[(Char, Int)] = occ match {
    case 0 => Nil
    case n => (char, n) :: combinate(char, n - 1)
  }
  def cartesianProd[T](l: List[List[T]]):List[List[T]] = l match {
    case Nil => List(List())
    case l1 :: tail => {
      val pr = cartesianProd(tail)
      l1.flatMap((x: T) => pr.map((l: List[T]) => x :: l)) ::: pr
    }
  }
  /** 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 = {
    x.flatMap((tx: (Char, Int)) => {
      y.find((ty: (Char, Int)) => tx._1 == ty._1) match {
        case None => List(tx)
        case Some(ty: (Char, Int)) => if (ty._2 == tx._2) Nil else List((tx._1, tx._2 - ty._2))
      }
    })
  }

  /** 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 sentenceAnagramOcc(occ: Occurrences): List[Sentence] = {
      val comb = combinations(occ)
      comb.flatMap((o: Occurrences) => {
        val dict = dictionaryByOccurrences(o)
        dict match {
          case Nil => List(List())
          case dict => dict.flatMap((w: Word) => {
            val subt = subtract(occ, o)
            val s = sentenceAnagramOcc(subt)
            val a = s.map((s: Sentence) => w :: s)
            if (s.forall((s: Sentence) => s == Nil) && subt != Nil) {
              println("FRFSGGSDGFLHUYSGFKCLVGSJYGFKLSDBGLKVHSIUDGVLKSFGVLKGSLKJVGBRSFL\nKGVSLDKHGFLKSGHFLKJHSDKLJHFLKJSDHFHKLSDHFLKJSDHFLKJSHDFJKHSDLFGHJSD")
              Nil
            } else a
            })
        }
      })
    }
    sentenceAnagramOcc(sentenceOccurrences(sentence))
  }
}