regex pattern to match string in any order

Question

How to find first match of string abc in shuffled order using regex pattern in java?

Example:

input 1: abcbaa
input 2: bcbaaa
input 3: cbaaab

1st match for input 1 : abcbaab
1st match for input 2 : bcbaaab
1st match for input 3 : bcaaabc

Patterns that I've tried that didn't work:

(?:([abc])(?!\\.*]\\1)){3}
(?!(.)\\1)[abc]{3}

The above 2 patterns matches 3 consecutive characters, including duplicate values.
example: ababac
expected: ababac

(?=.*[abc])(?=.*[abc])(?=.*[abc])

This one matches and empty character in-between each character. i.e., string position (0,0), (1,1), (2,2) etc...

Please show some code. What have you tried so far? SO will not build solutions, but answer specific questions. — Jonathan Lam, Jul 25 '16 at 13:08
Go read some tutorials about regex, you will find what you need. — Ephi, Jul 25 '16 at 13:09
Possible duplicate of [Java Regex looking a combination of words in any order](http://stackoverflow.com/questions/26041258/java-regex-looking-a-combination-of-words-in-any-order) — Alex K., Jul 25 '16 at 13:14
You might want to have a deeper look in the classes java.util.regex.Pattern and java.util.regex.Matcher — markus_, Jul 25 '16 at 13:14
Something like [`([abc])(?!\1)([abc])(?!\1|\2)[abc]`](https://regex101.com/r/mO8lG5/1) — Wiktor Stribiżew, Jul 25 '16 at 14:11
Thanks @WiktorStribiżew ! pattern you mentioned was very useful and helped me in improving performance too. Thanks again. — Arun, Jul 26 '16 at 04:42
@Arun Instead of putting @ Others, you'd better write comments to everyone who posted comments so that they could reopen your question if they think your question qualifies. The patterns are not enough as you should state what problems you have when using them. — Wiktor Stribiżew, Jul 26 '16 at 06:02

score 1 · Answer 1 · edited Jul 25 '16 at 13:19

1

Have you tried to look of all the possibilities? Something like this with your example:

(abc|acb|bca|bac|cab|cba)

edited Jul 25 '16 at 13:19

RealSkeptic

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answered Jul 25 '16 at 13:13

Reda Bourial

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This solution event if it is simple doesn't really give you the best performance (factorial(n) complexity) – Reda Bourial Jul 25 '16 at 13:16

The Anh Nguyen · Answer 2 · 2016-11-07T01:27:29.650

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Maybe you should try this regex:

^(?=[\s\S]*(a)+)(?=[\s\S]*(b)+)(?=[\s\S]*(c)+)[\s\S]*$

edited Nov 07 '16 at 01:27

answered Nov 04 '16 at 05:39

The Anh Nguyen

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`(.|\s)*` is [extremely inefficient](http://stackoverflow.com/questions/2407870/javascript-regex-hangs-using-v8) and should never be used. Instead, use Java's `DOTALL` flag, or add `(?s)` to the beginning of the regex. – Alan Moore Nov 04 '16 at 08:40
Thank you, Alan. Can I use [^]* instead of (.|\s)* ? – The Anh Nguyen Nov 04 '16 at 09:57
`[^]` is only useful in the JavaScript flavor, which doesn't have a DOTALL option (yet). In Java it's a syntax error. If you are using JavaScript, I recommend you use `[\s\S]` instead. It's longer, but it works the same across flavors, and its meaning is more obvious (any whitespace character + any non-whitespace character = any character). – Alan Moore Nov 04 '16 at 21:10

score 0 · Answer 3 · answered Nov 04 '16 at 08:35

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(?:a()|b()|c()){3}\1\2\3

The empty groups act like check boxes, so if \1\2\3 matches, each of the letters must have been seen at least once. Since the regex only consumes three characters, you know there's exactly one of each letter.

answered Nov 04 '16 at 08:35

Alan Moore

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regex pattern to match string in any order

3 Answers3