scala - Transform this iterative solution to functional solution -

this class takes map of type [string , list[string]] , outputs map of [string, string] key name of list , values binary representation of letters. each digit corresponds if letter appears in list or not. 1 - appears, 0 - not appear. example list :

1 = 1,1,0,0 2 = 1,1,1,0 3 = 1,1,0,1 4 = 1,1,0,0   returns  4-->1100 1-->1100 2-->1110 3-->1101 

below iterative solution :

object binaryrep {    var userdetails : scala.collection.immutable.hashmap[string, list[string]] = new  scala.collection.immutable.hashmap[string, list[string]]    var letterstocheck = list("a" , "b" , "c"  ,"d")   def main(args: array[string]) {      userdetails += "1" -> list("a" , "b")     userdetails += "2" -> list("a" , "b" , "c")     userdetails += "3" -> list("a" , "b" , "d")     userdetails += "4" -> list("a" , "b")      val binrep = getbinaryrepresentation      getbinaryrepresentation foreach ( (t2) => println (t2._1 + "-->" + t2._2))    }    def getbinaryrepresentation = {      var mapvalues = new scala.collection.immutable.hashmap[string, string]     var binaryrep = "";      (usd <- userdetails) {       (letter <- letterstocheck) {         if (usd._2.contains(letter)) {           binaryrep += "1"         } else {           binaryrep += "0";         }       }       mapvalues += usd._1 -> binaryrep       binaryrep = "";     }     mapvalues    }   } 

i think quite messy best do. more functional approach accomplish same result ?

import scala.collection.immutable.hashmap  object binaryrep {   val userdetails = hashmap("1" -> "ab",                             "2" -> "abc",                             "3" -> "abd",                             "4" -> "ab")   val letterstocheck = "abcd"   def getbinaryrepresentation = userdetails.mapvalues(string => letterstocheck.map(letter => if (string.contains(letter)) '1' else '0'))    getbinaryrepresentation foreach ( (t2) => println (t2._1 + "-->" + t2._2)) } 

data transformation can implement series of map calls.