How to separate a string by perens without losing the perens? Ruby

Question

So I'm trying to solve a challenge where if the perens are in the right order, I return true; else, of course, false. Code I came up with is here:

def valid_parentheses(str)
    return false if str.length % 2 == 1
        begin
            eval(str)
        rescue SyntaxError
            false
        else
            true
        end
    end
end

Works like a charm except in situations like valid_parentheses("hi(hi)()") which should returns # => true but instead returns false because the ending () are quote-unquote unnecessary and therefore raises an error.

I tried to split it out by parentheses, but:

str.split(/\(.*\))
# =>"hi"

because it deleted all the parens, and:

str.scan(/\(.*\))
#=> "(hi)()"

because it still technically starts with ( and ends with ).

How do I split that up to get "(hi)" and "()" separately?

Could you explain clearly what you want to achieve with one or several example input strings and the corresponding expected results? If I understand well, you want to check if parenthesis are balanced, isn't it? I can say already that the method `split` isn't the way to go. And why this strange test `return false if str.length % 2 == 1`? *(note that `== 1` is useless, and instead of using the modulo, you can check the last bit)* — Casimir et Hippolyte, Feb 27 '16 at 23:09

score 0 · Answer 1 · edited May 23 '17 at 12:07

You are close to what you want. Here is the regex I'd use if I understood correctly your question :

(\(.*?\))

\(      // match the left bracket
.*?     // match any character as FEW times as possible (lazy use of * because of the ?)
\)      // match the right bracket

Regex Demo

Indeed, thanks to the lazy mode, the search of the given character(s) will begin right after at the last consumed character.

X* start the search at end of the line (greedy)
X*? start the search right after the last consumed character (lazy)
X*+ is like the greedy quantifier except that it doesn't backtrack

@Anomie explain it here : Greedy vs. Reluctant vs. Possessive Quantifiers

Cary Swoveland · Accepted Answer · 2016-02-29T06:15:16.813

There are a few ways to do that. One is to first remove all characters other than "(" and ")", then in a loop remove substrings "()", continuing until no more such pairs can be removed. If the string is then empty, the parens are balanced; else they are not.

def balanced_parens?(str)
  pstr = str.gsub(/[^()]/,"")
  puts "pstr=#{pstr}"
  while pstr.gsub!("()", "")
  end
  pstr.empty?
end

balanced_parens? "Now (is(the(time)to(have(fun)))fun)"
  #=> true
balanced_parens? "Now (is(the)time)to(have)fun)((fun)"
  #=> false

Here are the step-by-step's for these two strings:

str = "Now (is(the(time)to(have(fun)))fun)"
pstr = str.gsub(/[^()]/,"")
  #=> "((()(())))" 
pstr.gsub!("()", "")
  #=> "((()))" 
pstr.gsub!("()", "")
  #=> "(())" 
pstr.gsub!("()", "")
  #=> "()" 
pstr.gsub!("()", "")
  #=> "" 
pstr.gsub!("()", "")
  #=> nil 
pstr.empty?
  #=> true

str = "Now (is(the)time)to(have)fun)((fun)"
pstr = str.gsub(/[^()]/,"")
  #=> "(())())(()" 
pstr.gsub!("()", "")
  #=> "())(" 
pstr.gsub!("()", "")
  #=> ")(" 
pstr.gsub!("()", "")
  #=> nil 
pstr.empty?
  #=> false

score -1 · Answer 3 · answered Feb 27 '16 at 22:16

-1

Have you considered using a regular expression? Something like:

/hay/ =~ 'haystack'   #=> 0
/y/.match('haystack') #=> #<MatchData "y">

might work.

answered Feb 27 '16 at 22:16

vibration

45
1
5

Just tried, actually. If I don't account for other characters (e.g. `/\(\)/.match(str)` ) it gives me "()", but if I do like I showed above, I still get "(hi)()". Thanks tho – Andrea McKenzie Feb 27 '16 at 22:19

How to separate a string by perens without losing the perens? Ruby

3 Answers3

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