Turning iteration into recursion

Question

I want to check if the string user entered has a balanced amount of ( and )\'s

ex. ()( is not balanced (())

Answer 1

A direct, equivalent conversion of the algorithm would look like this:

def check(string, counter=0):
  if not string:
    return "Balanced" if counter == 0 else "Unbalanced"
  elif counter < 0:
    return "Unbalanced"
  elif string[0] == "(":
    return check(string[1:], counter+1)
  elif string[0] == ")":
    return check(string[1:], counter-1)
  else:
    return check(string[1:], counter)

Use it like this:

check("(())")
=> "Balanced"

check(")(")
=> "Unbalanced"

Notice that the above algorithm takes into account cases where the closing parenthesis appears before the corresponding opening parenthesis, thanks to the elif counter < 0 condition - hence fixing a problem that was present in the original code.