Javascript removes letters duplicates and sequence in string

Question

There are many posts like this and I have found a few solutions but there are not perfect. One of them:

"aabbhahahahahahahahahahahasetsetset".replace(/[^\w\s]|(.+)\1+/gi, '$1')

The results is:

abhahahahahahaset

I want to get result:

abhaset

How to do this ?

Our answers fix your specific problem. But I wonder, what is your desired output for abcabcDabcabcD. Do you want abcDabcD or abcabcD? So longest, or shortest repetition if there are nested repetitions. — Martin Ender
– Martin Ender, Commented Apr 25, 2013 at 13:09
Ah okay, then just run the replacement until it doesn't change any more. — Martin Ender
– Martin Ender, Commented Apr 25, 2013 at 13:16

Martin Ender · Accepted Answer · 2013-04-25 13:22:41Z

4

.+ is greedy. It takes as much as it can. That is half of the has so that \1 can match the second half. Making the repetition ungreedy should do the trick:

/[^\w\s]|(.+?)\1+/gi

By the way, the i doesn't change anything here.

To get rid of nested repetitions (e.g. transform aaBBaaBB into aB (via aaBB or aBaB)) simply run the replacement multiple times until the result does not change any more.

var pattern = /[^\w\s]|(.+?)\1+/g;

var output = "aaBBaaBB";
var input;

do
{
    input = output;
    output = input.replace(pattern, "$1");
} while (input != output)

I admit the naming of output is a bit awkward for the first repetition, but you know... the two most difficult problems in computer science are cache invalidation, naming things and off-by-one errors.

edited Apr 25, 2013 at 13:22

answered Apr 25, 2013 at 13:07

Martin Ender

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Explosion Pills · Accepted Answer · 2013-04-25 13:09:13Z

2

.+ will match the maximum amount possible, so hahahaha satisfies (.+)\1 with haha and haha. You want to match the minimum amount possible, so use a reluctant quantifier.

"aabbhahahahahahahahahahahasetsetset".replace(/[^\w\s]|(.+?)\1+/gi, '$1')

http://jsfiddle.net/HQRDg/

answered Apr 25, 2013 at 13:09

Explosion Pills

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1 Comment

mitch Over a year ago

Thank you very much, but m.buettner was first :)

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