Creating white space in a List in Scala - scala

Question

I am trying to convert a List of string to the form "rBrrBB", or "r r rb", or " rrB". The string must have length 6. If the this list is not full, then the list should be prefixed with an appropriate number of spaces
So far my code as below
def showColumn(xs: List[String]): String = xs match
{case List() => ""
case x::xs1 => x.format(" ") + showColumn(xs1)}
when I call this from
println (showColumn(List("","","b","b","r","b")))
it returns only "bbrb". It suppose to return " bbrb"
Any help would appreciate.

Answer 1

Try this:
def showColumn(xs: List[String]): String = xs.map(x => if(x == "") " " else x).mkString
or, alternatively:
def showColumn(xs: List[String]): String = xs.map(x => if(x.isEmpty) " " else x).mkString
Both work by changing empty strings in the list to spaces, then merging each string in the list into a single string.
If you absolutely must make this a recursive function, then a solution which is not tail-recursive would look like this:
def showColumn(xs: List[String]): String = xs match {
case Nil => ""
case x :: xs1 => (if(x.isEmpty) " " else x) + showColumn(xs1)
}
Finally, the tail-recursive version is a little more complex, as it employs a helper function:
import scala.annotation.tailrec
def showColumn(xs: List[String]): String = {
// Tail recursive helper function.
#tailrec
def nextStr(rem: List[String], acc: String): String = rem match {
case Nil => acc
case x :: xs1 => nextStr(xs1, acc + (if(x.isEmpty) " " else x))
}
// Start things off.
nextStr(xs, "")
}