How do I count the number of consecutive occurrences of a character in a string in Scala?

Question

Let's say I have the string:

"AAAAA BBB CCCC AA"

I want a way that I can count the occurrences of consecutive characters to give a string that looks like:

"5A 3B 4C 2A"

How do I go about this?

Currently, I have tried looping through the string, and for each character, I use a match case, and within each case, I start a new loop to count how many times that character appears consecutively, I then try to replace that substring with the desired substring.

Answer 1

Build a List of (n:Int, c:Char) tuples then reformat it to whatever String representation you desire.

"AAAAABBBCCCCAA".foldRight(List.empty[(Int,Char)]){
  case (c, hd::tl) if c == hd._2 => (hd._1 + 1, c) :: tl
  case (c, lst) => (1,c) :: lst
}.map(tup => s"${tup._1}${tup._2}").mkString(" ")
//res0: String = 5A 3B 4C 2A

Answer 2

For this kind of problems, I like to write my own tail-recursive algorithm.

def countConsecutiveCharacters(str: String): List[(Char, Int)] = {
  @annotation.tailrec
  def loop(remaining: List[Char], currentChar: Char, currentCount: Int,
           acc: List[(Char, Int)]): List[(Char, Int)] =
    remaining match {
      case char :: xs if(char == currentChar) =>
        loop(
          remaining = xs,
          currentChar,
          currentCount + 1,
          acc
        )

      case char :: xs =>
        loop(
          remaining = xs,
          currentChar = char,
          currentCount = 1,
          (currentChar -> currentCount) :: acc
        )

      case Nil =>
       ((currentChar -> currentCount) :: acc).reverse
    }

  str.toList match {
    case char :: list =>
      loop(
        remaining = list,
        currentChar = char,
        currentCount = 1,
        acc = List.empty
      )

    case Nil =>
      List.empty
  }
}

You can checkout the code working here .

You may replace all that with one foldLeft , but IMHO, this way is cleaner and easier to read.