How to balance parenthesis recursively

Question

I\'m working on some code to balance parenthesis, this question proved most useful for the algorithm.

I implemented it in my first language (PHP) but I\'m learning Scala

Answer 1

Adding to Vigneshwaran's answer (including comments & filtering unnecessary letters to avoid extra recursive calls):

def balance(chars: List[Char]): Boolean = {
  @scala.annotation.tailrec
  def recurs_balance(chars: List[Char], openings: Int): Boolean = {
    if (chars.isEmpty) openings == 0
    else if (chars.head == '(') recurs_balance(chars.tail, openings + 1)
    else openings > 0 && recurs_balance(chars.tail, openings - 1)
  }

  recurs_balance(chars.filter(x => x == '(' || x == ')'), 0)
}