Scala: Make sure braces are balanced

Question

I am running a code to balance brackets in statement. I think i have gotten it correct but it is failing on one particular statement, i need to understand why?

This

Answer 1

You can also use the property of Stack data structure to solve this problem. When you see open bracket, you push it into the stack. When you see close bracket, you pop from the stack (instead of Stack I'm using List, because immutable Stack is deprecated in Scala):

def isBalanced(chars: Seq[Char]): Boolean = {
  import scala.annotation.tailrec

  case class BracketInfo(c: Char, idx: Int)
  def isOpen(c: Char): Boolean = c == '('
  def isClose(c: Char): Boolean = c == ')'
  def safePop[T](stack: List[T]): Option[T] = {
    if (stack.length <= 1) stack.headOption
    else stack.tail.headOption
  }

  @tailrec
  def isBalanced(chars: Seq[Char], idx: Int, stack: List[BracketInfo]): Boolean = {
    chars match {
      case Seq(c, tail@_*) =>
        val newStack = BracketInfo(c, idx) :: stack // Stack.push
        if (isOpen(c)) isBalanced(tail, idx + 1, newStack)
        else if (isClose(c)) {
          safePop(stack) match {
            case Some(b) => isBalanced(tail, idx + 1, stack.tail)
            case None =>
              println(s"Closed bracket '$c' at index $idx was not opened")
              false
          }
        }
        else isBalanced(tail, idx + 1, stack)
      case Seq() =>
        if (stack.nonEmpty) {
          println("Stack is not empty => there are non-closed brackets at positions: ")
          println(s"${stack.map(_.idx).mkString(" ")}")
        }
        stack.isEmpty
    }
  }

  isBalanced(chars, 0, List.empty[BracketInfo])
}