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

  val myStack = new Stack[Char]

  def balance(chars: List[Char]): Boolean = {
    def processParanthesis(x: Char, a: List[Char]): Stack[Char] = {
      if (x == '(') {
        myStack.push('(');
      } else if (x == ')') {
        if (!myStack.empty())
          myStack.pop();
        else
          myStack.push(')');
      }
      if (a.length == 0)
        return myStack;
      else
        return processParanthesis(a.head, a.tail);
    }
    return processParanthesis(chars.head, chars.tail).empty();
  }

Answer 2

My solution for this

def balance(chars: List[Char]): Boolean = {
 var braceStack = new Stack[Char]()

def scanItems(strList:List[Char]):Boolean = {
   if(strList.isEmpty)
       braceStack.isEmpty
   else{
      var item = strList.head
      item match {
        case '(' => braceStack.push(item)
                    scanItems(strList.tail)
        case ')'=> if(braceStack.isEmpty){
                        false
                    }
                    else {
                      braceStack.pop
                      scanItems(strList.tail)
                    }
        case _ => scanItems(strList.tail)
      }
    }
 }

 scanItems(chars)

}

Answer 3

Instead of using Switch case you can use recursion to solve your problem in an efficient way.

Following is my code to achieve the same with the help of recursion. You need to convert the string to List for my method.

Code :

object Balance {

  def main(args: Array[String]): Unit = {
    var paranthesis = "(234(3(2)s)d)" // Use your string here
    println(bal(paranthesis.toList)) // converting the string to List 
  }

  def bal(chars: List[Char]): Boolean ={
   // var check  = 0
    def fun(chars: List[Char],numOfOpenParan: Int): Boolean = {
      if(chars.isEmpty){
        numOfOpenParan == 0
      }
      else{
        val h = chars.head

        val n = 
          if(h == '(') numOfOpenParan + 1
          else if (h == ')') numOfOpenParan - 1
          else numOfOpenParan 

       // check  = check + n
        if (n >= 0) fun(chars.tail,n)
        else false 
      }
    }
    fun(chars,0)
  }
}

Answer 4

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)
}

Answer 5

It seems we are attending the same course. my solution:

def balance(chars: List[Char]): Boolean = 
doBalance(chars, 0) == 0;
def doBalance(chars: List[Char], parenthesisOpenCount: Int): Int =
if(parenthesisOpenCount <0) -100;
else
if(chars.isEmpty) parenthesisOpenCount
else
  chars.head match {
  case '(' => return doBalance(chars.tail, parenthesisOpenCount+1) 
  case ')' => return doBalance(chars.tail, parenthesisOpenCount-1)
  case _ => return doBalance(chars.tail, parenthesisOpenCount)
}

Answer 6

Same as Aaron Novstrup's answer but using 'if else'. I'm also attending the same course but we are taught upto if/else only so far.

def balance(chars: List[Char]): Boolean = {
    def balanced(chars: List[Char], open: Int): Boolean = 
      if (chars.isEmpty) open == 0
      else if (chars.head == '(') balanced(chars.tail, open + 1)
      else if (chars.head == ')') open > 0 && balanced(chars.tail, open - 1)
      else balanced(chars.tail, open)
    balanced(chars, 0)
}