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Pattern matching : Matching a String with a pattern

I was trying to match a pattern within a string. I am running out of ideas how to do this in Java with good time complexity.
No its not a simple regex matching (but loved to be proved wrong)
What I am trying is,
Pattern : "1221" (Means 1 word repeat once, 2nd word repeat twice, last word is same as first word)
Valid Input: "aabbbbbbaa" (aa occurs at the beginning and at end, while middle portion is occupied by bbb repeating twice)
I tried the following approaches but failed miserably
I tried to loop input with the pattern. But that did not solve the problem, although with more loops I can achieve it, but it increases the time complexity exponentially.
Tried recursion and again no use.
What other approaches can I try?
I think Dynamic programming might be the answer, but I am not able to determine the terminating condition.
Any help would be appreciated.

You can use simple regex, e.g.:
^(.+)(.+)\2\1$
It does exactly what u want:

Java, poor regex performance with lazy expressions

The code is actually in Scala (Spark/Scala) but the library scala.util.matching.Regex, as per the documentation, delegates to java.util.regex.
The code, essentially, reads a bunch of regex from a config file and then matches them against logs fed to the Spark/Scala app. Everything worked fine until I added a regex to extract strings separated by tabs where the tab has been flattened to "#011" (by rsyslog). Since the strings can have white-spaces, my regex looks like:
(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)#011(.+?)
The moment I add this regex to the list, the app takes forever to finish processing logs. To give you an idea of the magnitude of delay, a typical batch of a million lines takes less than 5 seconds to match/extract on my Spark cluster. If I add the expression above, a batch takes an hour!
In my code, I have tried a couple of ways to match regex:
if ( (regex findFirstIn log).nonEmpty ) { do something }
val allGroups = regex.findAllIn(log).matchData.toList
if (allGroups.nonEmpty) { do something }
if (regex.pattern.matcher(log).matches()){do something}
All three suffer from poor performance when the regex mentioned above it added to the list of regex. Any suggestions to improve regex performance or change the regex itself?
The Q/A that's marked as duplicate has a link that I find hard to follow. It might be easier to follow the text if the referenced software, regexbuddy, was free or at least worked on Mac.
I tried negative lookahead but I can't figure out how to negate a string. Instead of /(.+?)#011/, something like /([^#011]+)/ but that just says negate "#" or "0" or "1". How do I negate "#011"? Even after that, I am not sure if negation will fix my performance issue.

The simplest way would be to split on #011. If you want a regex, you can indeed negate the string, but that's complicated. I'd go for an atomic group
(?>(.+?)#011)
Once matched, there's no more backtracking. Done and looking forward for the next group.
Negating a string
The complement of #011 is anything not starting with a #, or starting with a # and not followed by a 0, or starting with the two and not followed... you know. I added some blanks for readability:
((?: [^#] | #[^0] | #0[^1] | #01[^1] )+) #011
Pretty terrible, isn't it? Unlike your original expression it matches newlines (you weren't specific about them).
An alternative is to use negative lookahead: (?!#011) matches iff the following chars are not #011, but doesn't eat anything, so we use a . to eat a single char:
((?: (?!#011). )+)#011
It's all pretty complicated and most probably less performant than simply using the atomic group.
Optimizations
Out of my above regexes, the first one is best. However, as Casimir et Hippolyte wrote, there's a room for improvements (factor 1.8)
( [^#]*+ (?: #(?!011) [^#]* )*+ ) #011
It's not as complicated as it looks. First match any number (including zero) of non-# atomically (the trailing +). Then match a # not followed by 011 and again any number of non-#. Repeat the last sentence any number of times.
A small problem with it is that it matches an empty sequence as well and I can't see an easy way to fix it.

Java & Regex: Matching a substring that is not preceded by specific characters

This is one of those questions that has been asked and answered hundreds of times over, but I'm having a hard time adapting other solutions to my needs.
In my Java-application I have a method for censoring bad words in chat messages. It works for most of my words, but there is one particular (and popular) curse word that I can't seem to get rid of. The word is "faen" (which is simply a modern slang for "satan", in the language in question).
Using the pattern "fa+e+n" for matching multiple A's and E's actually works; however, in this language, the word for "that couch" or "that sofa" is "sofaen". I've tried a lot of different approaches, using variations of [^so] and (?!=so), but so far I haven't been able to find a way to match one and not the other.
The real goal here, is to be able to match the bad words, regardless of the number of vowels, and regardless of any non-letters in between the components of the word.
Here's a few examples of what I'm trying to do:
"String containing faen" Should match
"String containing sofaen" Should not match
"Non-letter-censored string with f-a#a-e.n" Should match
"Non-letter-censored string with sof-a#a-e.n" Should not match
Any tips to set me off in the right direction on this?

You want something like \bf[^\s]+a[^\s]+e[^\s]+n[^\s]\b. Note that this is the regular expression; if you want the Java then you need to use \\b[^\\s]+f[^\\s]+a[^\\s]+e[^\\s]+n[^\\s]\b.
Note also that this isn't perfect, but does handle the situations that you have suggested.

It's a terrible idea to begin with. You think, your users would write something like "f-aeen" to avoid your filter but would not come up with "ffaen" or "-faen" or whatever variation that you did not prepare for? This is a race you cannot win and the real loser is usability.

Java Regex Engine Crashing

Regex Pattern - ([^=](\\s*[\\w-.]*)*$)
Test String - paginationInput.entriesPerPage=5
Java Regex Engine Crashing / Taking Ages (> 2mins) finding a match. This is not the case for the following test inputs:
paginationInput=5
paginationInput.entries=5
My requirement is to get hold of the String on the right-hand side of = and replace it with something. The above pattern is doing it fine except for the input mentioned above.
I want to understand why the error and how can I optimize the Regex for my requirement so as to avoid other peculiar cases.

You can use a look behind to make sure your string starts at the character after the =:
(?<=\\=)([\\s\\w\\-.]*)$
As for why it is crashing, it's the second * around the group. I'm not sure why you need that, since that sounds like you are asking for :
A single character, anything but equals
Then 0 or more repeats of the following group:
Any amount of white space
Then any amount of word characters, dash, or dot
End of string
Anyway, take out that *, and it doesn't spin forever anymore, but I'd still go for the more specific regex using the look behind.
Also, I don't know how you are using this, but why did you have the $ in there? Then you can only match the last one in the string (if you have more than one). It seems like you'd be better off with a look-ahead to the new line or the end: (?=\\n|$)
[Edit]: Update per comment below.

Try this:
=\\s*(.*)$

Best practices for regex performance VS sheer iteration

I was wondering if there are any general guidelines for when to use regex VS "string".contains("anotherString") and/or other String API calls?
While above given decision for .contains() is trivial (why bother with regex if you can do this in a single call), real life brings more complex choices to make. For example, is it better to do two .contains() calls or a single regex?
My rule of thumb was to always use regex, unless this can be replaced with a single API call. This prevents code against bloating, but is probably not so good from code readability point of view, especially if regex tends to get big.
Another, often overlooked, argument is performance. How do I know how many iterations (as in "Big O") does this regex require? Would it be faster than sheer iteration? Somehow everybody assumes that once regex looks shorter than 5 if statements, it must be quicker. But is this always the case? This is especially relevant if regex cannot be pre-compiled in advance.

RegexBuddy has a built-in regular expressions debugger. It shows how many steps the regular expression engine needed to find a match or to fail to find a match. By using the debugger on strings of different lengths, you can get an idea of the complexity (big O) of the regular expression. If you look up "benchmark" in the index of RegexBuddy's help file you'll get some more tips on how to interpret this.
When judging the performance of a regular expression, it is particularly important to test situations where the regex fails to find a match. It is very easy to write a regular expression that finds its matches in linear time, but fails in exponential time in a situation that I call catastrophic backtracking.
To use your 5 if statements as an example, the regex one|two|three|four|five scans the input string once, doing a little bit of extra work when an o, t, or f is encountered. But 5 if statements checking if the string contains a word will search the entire string 5 times if none of the words can be found. If five occurs at the start of the string, then the regex finds the match instantly, while the first 4 if statements scan the whole string in vain before the 5th if statement finds the match.

It's hard to estimate performance without using a profiler, generally the best strategy is to write what makes the most logical sense and is easier to understand/read. If two .contains() calls are easier to logically understand then that's the better route, the same logic applies if a regex makes more sense.
It's also important to consider that other developers on your team may not have a great understanding of regex. If at a later time in production the use of regex over .contains() (or vice versa) is identified as a bottleneck, try and profile both.
Rule of thumb: Write code to be readable, use a profiler to identify bottlenecks and only then replace the readable code with faster code.

I would strongly suggest that you write the code for both and time it. It's pretty simple to do this and you'll get an answers that is not a generic "rule of thumb" but instead a very specific answer that holds for your problem domain.
Vance Morrison has an excellent post about micro benchmarking, and has a tool that makes it really simple for you to answer questions like this...
http://msdn.microsoft.com/en-us/magazine/cc500596.aspx
If you want my personal "rule of thumb" then it's that RegEx is often slower for this sort of thing, but you should ignore me and measure it yourself :-)
If, for non-performance reasons, you continue to use Regular Expressions then I can really recommend two things. Get a profiler (such as ANTS) and see what your code does in production. Then, get a copy of the Regular Expression Cookbook...
http://www.amazon.co.uk/Regular-Expressions-Cookbook-Jan-Goyvaerts/dp/0596520689/ref=sr_1_1?ie=UTF8&s=books&qid=1259147763&sr=8-1
... as it has loads of tips on speeding up RegEx code. I've optimized RegEx code by a factor of 10 following tips from this book.

The answer (as usual) is that it depends.
In your particular case, I guess the alternative would be to do the regex "this|that" and then do a find. This particular construct really pokes at regex's weaknesses. The "OR" in this case doesn't really know what the sub-patterns are trying to do and so can't easily optimize. It ends up doing the equivalent of (in pseudo code):
for( i = 0; i < stringLength; i++ ) {
if( stringAt pos i starts with "this" )
found!
if( stringAt pos i starts with "that" )
found!
}
There almost isn't a slower way to do it. In this case, two contains() calls will be much faster.
On the other hand, a full match on: ".*this.*|.*that.*" may optimize better.
To me, regex should be used when the code to do otherwise is complicated or unwieldy. So if you want to find one of two or three strings in a target string then just use contains. But if you wanted to find words starting with 'A' or 'B' and ending in 'g'-'m'... then use regex.
And then you won't be so worried about a few cycles here and there.