I have a strings like:
0011
01
000111
000111
I need to validate them like this: count of "0" must be identical to the count of "1". So "001" - invalid, "0011" - valid.
How can I do this with regex?
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What language are you using? – Wiktor Stribiżew May 13 '15 at 11:11
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1Check the answers of this question: http://stackoverflow.com/questions/3644266/how-can-we-match-an-bn-with-java-regex – axiac May 13 '15 at 11:17
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1Or check the string against `/(0+)(1+)/` then compare the lengths of the captured sub-expressions. – axiac May 13 '15 at 11:21
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Actually, the question mentioned does not have an example for Ruby. – Wiktor Stribiżew May 13 '15 at 21:34
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@stribizhev Care to be more explicit about why those regexes won't work with Ruby's regex engine? – mu is too short May 13 '15 at 22:01
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@muistooshort: In Ruby, the subroutines have different syntax that is rather misleading since it looks like a back-reference. – Wiktor Stribiżew May 13 '15 at 22:07
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In Ruby, you can use subroutines:
m = /\b(0(\g<1>)?1)\b/.match('000111');
puts m;
000111
Or, you can just use capturing groups to match adjacent 0s and 1s, and then check the captured group length:
m = /(0+)(1+)/.match('0011');
puts m[1].length === m[2].length ? "True" : "False";
m = /(0+)(1+)/.match('00111');
puts m[1].length === m[2].length ? "True" : "False";
You may add ^ and $ to only match a string consisting of leading zeros and trailing 1s (m = /^(0+)(1+)$/.match('00111');).
Output of the demo program:
True
False
Wiktor Stribiżew
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It can be done if you adapt `'/ ^ (a (?1)? b) $ /x'` to Ruby: http://rubular.com/r/9JZfVgVosj – nhahtdh May 13 '15 at 12:10
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I have found out where I was wrong thanks to [rexegg.com](http://www.rexegg.com/regex-disambiguation.html) I was just unaware of the `\g<1>` notation. Your suggestion - `^(?
a\g – Wiktor Stribiżew May 13 '15 at 12:15?b)$` - is also working well, as I see. -