Discussion on: Daily Challenge #151 - Reverse Parentheses

#!/usr/bin/env python3


def reverseInParens(string):
    """Reverse a given string as per dev.to daily challenge #151 
    """
    # the basic idea is simple: keep a stack of strings representing the paren-delimited
    # parts of our input - for instance, given the string h(le)lo, by the time we parse e,
    # our stack looks like:
    # 1 - "el"
    # 0 - "h"
    # every time we encounter an lparen, we add a new entry into the stack, and
    # every time we encounter an rparen, we merge the newest entry on the stack with the
    # one directly behind it with a pair of parens, so going back to our previous example,
    # after parsing the rparen, our stack looks like this:
    # 0 - h(el)
    # adding characters is as simple as recognizing whether we should add in normal or
    # reverse order to the newest entry in the stack, which is as simple as seeing if
    # the index of the newest entry is odd - if it is, we add in reverse order, whereas
    # if it's odd, we add in normal order
    output_stack = [""]
    stack_index = 0
    for ch in string:
        if ch == "(":
            # add an empty string to our stack
            stack_index += 1
            output_stack.append("")
        elif ch == ")":
            # pop the last-added string and merge it with the one prior
            current_string = output_stack.pop()
            stack_index -= 1
            built_string = "(" + current_string + ")"
            # the below line did not take into account at which end of the prior string
            # our last-added string should be merged with
            # output_stack[stack_index] += "(" + current_string + ")"
            if stack_index % 2 == 1:
                output_stack[stack_index] = built_string + output_stack[stack_index]
            else:
                output_stack[stack_index] += built_string
        else:
            if stack_index % 2 == 1:
                # add in reverse order if we're on an odd number of parens, as
                # per the challenge
                output_stack[stack_index] = ch + output_stack[stack_index]
            else:
                # we're on an even number of parens/no parens, add the character
                # in non-reversed order (i.e, the order that the input string
                # already has)
                output_stack[stack_index] += ch
    # the final finished string is at index 0 of output_stack
    return output_stack.pop()


if __name__ == "__main__":
    print(reverseInParens("h(el)lo"))
    print(reverseInParens("a ((d e) c b)"))
    print(reverseInParens("one (two (three) four)"))
    print(reverseInParens("one (ruof ((rht)ee) owt)"))