Scala solution to TopCoder exercise CmpdWords: http://community.topcoder.com/stat?c=problem_statement&pm=3490&rd=8001 #LearningScala

CmpdWords.scala

object CmpdWords {
    
  def permutedPairs(elems:List[String]): List[String] = {
    elems match {
  		case Nil => List()
  		case x::Nil => List()
  		case x::xs => {
  			for {x <- elems;
  					 y <- elems-x} yield x+y
  		}
    }
  }
  
  def frequency(words: List[String]) = words.foldLeft(Map[String,Int]())((map,word) => {
	    val count = map.getOrElse(word,0)+1;
	    map + ((word,count))
    })
  
  def ambiguous(dictionary: Array[String]) : Int = {
  	val compoundWords = permutedPairs(dictionary.toList)
  	val compoundWordFrequency = frequency(compoundWords)
    compoundWords.toSet.count(x => dictionary.contains(x) || compoundWordFrequency(x)>1)
  }

  ambiguous(Array("a","aa","aaa"))                             //> res0: Int = 3
  ambiguous(Array("am","eat","a", "meat", "hook","meathook"))  //> res1: Int = 2
  ambiguous(Array("abc","bca","bab","a"))                     //> res2: Int = 1
}

Nice! You can simplify how to compute all the combinations of two different words in a list like this:

def permutedPairs(elems:List[String]): List[String] = 
  for {
    x <- elems
    y <- elems if y != x
  } yield x+y