Ex 2.2

package dk.itu.BIDMT.F19.P1.Part2

object SortingLists {

  /**
    *
    * Sort the input listOfLists in an ascending order  based on the length of each sub-list and return the sorted list
    *
    * For example, the sorting of the list: List(List('a', 'b', 'c'), List('d', 'e'), List('f', 'g', 'h'), List('d', 'e'), List('i', 'j', 'k', 'l'), List('m', 'n'), List('o'))
    * should be: List(List(o), List(d, e), List(d, e), List(m, n), List(a, b, c), List(f, g, h), List(i, j, k, l))
    */
  def sortListLength(listOfLists: List[List[Char]]): List[List[Char]]  =
    listOfLists sortBy {_.length}

  /**
    * Sort the sublists of the input list in an ascending order  based on the frequency of the lengths of these sublists.
    *
    * For example, the sorting of the list: List(List('a', 'b', 'c'), List('d', 'e'), List('f', 'g', 'h'), List('d', 'e'), List('i', 'j', 'k', 'l'), List('m', 'n'), List('o'))
    * should be: List(List(i, j, k, l), List(o), List(a, b, c), List(f, g, h), List(d, e), List(d, e), List(m, n))
    *
    * since:
    *   freq of sub-lists of length 4 is 1
    *   freq of sub-lists of length 1 is 1
    *   freq of sub-lists of length 3 is 2
    *   freq of sub-lists of length 2 is 3
    *
    *  Hint:
    *  Follow the following steps:
    *   - pair each sub-list with its length
    *   - find the freq of each length, you can use groupby for that
    *   - sort the groups according to the number of sub-lists in each group
    *   - generate the final sorted list
    */
  def sortListFreq(listOfLists: List[List[Char]]): List[List[Char]] = {
    val pairs = listOfLists.map(e => (e.length, e))
    val length_freq = pairs.groupBy(_._1).mapValues(_.map(_._2))
    length_freq.values.toList.sortBy(_.length).flatten
}


  def main(args: Array[String]):Unit = {
    val ls = List(List('a', 'b', 'c'), List('d', 'e'), List('f', 'g', 'h'), List('d', 'e'), List('i', 'j', 'k', 'l'), List('m', 'n'), List('o'))
    println("Sorted list according to lengths of sublists: "+ sortListLength(ls))
    println("Sorted list according to frequency of lengths of sublists: "+ sortListFreq(ls))
  }
}