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

You have mixed up the cases for ( and ).

Answer 2

Just for completeness, I found an even more terse 'scala-esque' implementation from another SO question:

def balance(chars: List[Char]): Boolean = chars.foldLeft(0){
  case (0, ')') => return false
  case (x, ')') => x - 1
  case (x, '(') => x + 1
  case (x, _  ) => x
} == 0

Answer 3

For what it's worth, here's a more idiomatic Scala implementation:

def balance(chars: List[Char]): Boolean = {
  @tailrec def balanced(chars: List[Char], open: Int): Boolean = 
    chars match {
      case      Nil => open == 0
      case '(' :: t => balanced(t, open + 1)
      case ')' :: t => open > 0 && balanced(t, open - 1)
      case   _ :: t => balanced(t, open)
    }

  balanced(chars, 0)
}