Remove unmatched parentheses from a string - ruby

Question

I want to remove "un-partnered" parentheses from a string.
I.e., all ('s should be removed unless they're followed by a ) somewhere in the string. Likewise, all )'s not preceded by a ( somewhere in the string should be removed.
Ideally the algorithm would take into account nesting as well.
E.g.:
"(a)".remove_unmatched_parents # => "(a)"
"a(".remove_unmatched_parents # => "a"
")a(".remove_unmatched_parents # => "a"

Answer 1

Instead of a regex, consider a push-down automata, perhaps. (I'm not sure if Ruby regular expressions can handle this, I believe Perl's can).
A (very trivialized) process may be:
For each character in the input string:
If it is not a '(' or ')' then just append it to the output
If it is a '(' increase a seen_parens counter and add it
If it is a ')' and seen_parens is > 0, add it and decrease seen_parens. Otherwise skip it.
At the end of the process, if seen_parens is > 0 then remove that many parens, starting from the end. (This step can be merged into the above process with use of a stack or recursion.)
The entire process is O(n), even if a relatively high overhead
Happy coding.

Answer 2

The following uses oniguruma. Oniguruma is the regex engine built in if you are using ruby1.9. If you are using ruby1.8, see this: oniguruma.
Update
I had been so lazy to just copy and paste someone else's regex. It seemed to have problem.
So now, I wrote my own. I believe it should work now.
class String
NonParenChar = /[^\(\)]/
def remove_unmatched_parens
self[/
(?:
(?<balanced>
\(
(?:\g<balanced>|#{NonParenChar})*
\)
)
|#{NonParenChar}
)+
/x]
end
end
(?<name>regex1) names the (sub)regex regex1 as name, and makes it possible to be called.
?g<name> will be a subregex that represents regex1. Note here that ?g<name> does not represent a particular string that matched regex1, but it represents regex1 itself. In fact, it is possible to embed ?g<name> within (?<name>...).
Update 2
This is simpler.
class String
def remove_unmatched_parens
self[/
(?<valid>
\(\g<valid>*\)
|[^()]
)+
/x]
end
end

Answer 3

Build a simple LR parser:
tokenize, token, stack = false, "", []
")(a))(()(asdf)(".each_char do |c|
case c
when '('
tokenize = true
token = c
when ')'
if tokenize
token << c
stack << token
end
tokenize = false
when /\w/
token << c if tokenize
end
end
result = stack.join
puts result
running yields:
wesbailey#feynman:~/code_katas> ruby test.rb
(a)()(asdf)
I don't agree with the folks modifying the String class because you should never open a standard class. Regexs are pretty brittle for parser and hard to support. I couldn't imagine coming back to the previous solutions 6 months for now and trying to remember what they were doing!

Answer 4

Here's my solution, based on #pst's algorithm:
class String
def remove_unmatched_parens
scanner = StringScanner.new(dup)
output = ''
paren_depth = 0
while char = scanner.get_byte
if char == "("
paren_depth += 1
output << char
elsif char == ")"
output << char and paren_depth -= 1 if paren_depth > 0
else
output << char
end
end
paren_depth.times{ output.reverse!.sub!('(', '').reverse! }
output
end
end

Answer 5

Algorithm:
Traverse through the given string.
While doing that, keep track of "(" positions in a stack.
If any ")" found, remove the top element from the stack.
If stack is empty, remove the ")" from the string.
In the end, we can have positions of unmatched braces, if any.
Java code:
Present # http://a2ajp.blogspot.in/2014/10/remove-unmatched-parenthesis-from-given.html