The PCRE-compatible regex to match this would be:

/([\s.',-])\1+/

If you're using Perl, you can replace it using the following expression:

s/([\s.',-])\1+/$1/g

If you're using PHP, then you would use this syntax:

$out = preg_replace('/([\s.\',-])\1+/', '$1', $in);

Explanation

• The () group matches the single character, in this case either a whitespace character (\s) or the punctuation characters (. ' - ,). It's good practice to put - at the end of the list inside [].
• The \1 means that the same thing it just matched in the parentheses occurs at least once more.
• In the replacement, the $1 refers to the match in first set of parentheses.

Note: this is Perl-Compatible Regular Expression (PCRE) syntax.

From the perlretut man page:

Matching repetitions

The examples in the previous section display an annoying weakness. We were only matching 3-letter words, or chunks of words of 4 letters or less. We'd like to be able to match words or, more generally, strings of any length, without writing out tedious alternatives like \w\w\w\w|\w\w\w|\w\w|\w.

This is exactly the problem the quantifier metacharacters ?, *, +, and {} were created for. They allow us to delimit the number of repeats for a portion of a regexp we consider to be a match. Quantifiers are put immediately after the character, character class, or grouping that we want to specify. They have the following meanings:

• a? means: match 'a' 1 or 0 times

• a* means: match 'a' 0 or more times, i.e., any number of times

• a+ means: match 'a' 1 or more times, i.e., at least once

• a{n,m} means: match at least "n" times, but not more than "m" times.

• a{n,} means: match at least "n" or more times

• a{n} means: match exactly "n" times

$ tr -s '[:lower:]' <<<"staaacksoveerfloow"
stacksoverflow

The tr utility is used here with its -s option to remove consecutive duplicates of any lowercase character in the given string.

In place of [:lower:], you could use a-z or any range or character class that matches the characters that you want to affect.

Change your regex like below.

string.replaceAll("((.)\\2{2})\\2+","$1");

• ( start of the first caturing group.
• (.) captures any character. For this case, you may use [a-z]
• \\2 refers the second capturing group. \\2{2} which must be repeated exactly two times.
• ) End of first capturing group. So this would capture the first three repeating characters.
• \\2+ repeats the second group one or more times.

DEMO

I think this is oddly complicated...though, I'm 99% sure this can be done with a single line of RegEx, I don't know how from the top of my head¹.

This can be shortened to a much simpler and shorter solution, without the need of arrays at all.

String input = "hhiii!!!  hooowww   aaareee   yyyooou???";
StringBuilder output = new StringBuilder();

// Append the first character of the string.
// We can also let the loop start at 0, but then
// we'd need an additional if inside the loop, which
// I'd like to avoid.
output.append(input.charAt(0));

// Start the loop at 1, because we already have the first character.
// We can not 
for (int idx = 1; idx < input.length(); idx++) {
    // Check the current against the previous character.
    if(input.charAt(idx) != input.charAt(idx-1)) {
        // If it is not the same, append it.
        output.append(input.charAt(idx));
    }
 }

Of course you'd need to protect yourself from invalid input (empty or null String), and it will also break words with allowed double-characters: foobar, tool, support etc..

Some further comments on your code:

• Function name: It's not directly visible what the function does. It should be called something along the lines of removeDoubleChars or removeMultipleCharacters.
• Function type: The function shouldn't directly output the result to System.out but rather return it.
• Loops: Use appropriate loops, f.e. your while should be a for.
• Variable names: I know it's partly taught, but I get the creeps (or become a Creeper) if I see variables which are named i, j or arr. Give your variables meaningful names! In any modern language the name of the variable is neither limited nor does it matter for performance. Write your code primarily for humans, and only secondary for machines...the coder which has to maintain your code after you will thank you.
• Whitespaces: Use 'em! if(str.charAt(i)!=str.charAt(i+1)){ is harder to read then if(str.charAt(i) != str.charAt(i + 1)) {.
• Code duplication: Avoid duplicate code. In this case it's minor, but the i++ should be moved outside the if, so that it's not necessary to have an else part.

¹: But John does.

You can run a python script within the PythonScript plugin.

If it is not yet installed, follow this guide

This will replace 2 or more spaces with 100 carriage return, as explained in your question.

• Create a script (Plugins >> PythonScript >> New Script)
• Copy this code and save the file (for example format.py):

import re
def format(match):
    return 100 * "\r"
editor.rereplace('\s\s+', format)

• Open the file you want to modify
• Run the script (Plugins >> PythonScript >> Scripts >> format)
• Done

Change it to:

string = string.replace(/-{2,}/g,"-");

Another way is

string = string.replace(/-+/g,"-");

as that replaces any one or more instances of - with only one -.

• Ctrl+H
• Find what: ^(\w+,\h*)(\w+)(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?(?:,\h*(\w+))?
• Replace with: $1$2\n(?3$1$3)(?4\n$1$4)(?5\n$1$5)(?6\n$1$6)(?7\n$1$7)(?8\n$1$8)(?9\n$1$9)
• TICK Match case
• TICK Wrap around
• SELECT Regular expression
• UNTICK . matches newline
• Replace all

Explanation:

^               # beginning of line
    (               # group 1
        \w+             # 1 or more word characters
        ,               # a comma
        \h*             # 0 or more horizontal spaces
    )               # end group 1
    (\w+)           # group 2, 1 or more word characters
    (?:             # non capture group
        \h*             # 0 or more horizontal spaces
        (\w+)           # group 3, 1 or more word characters
    )?              # end group, optional
(?:,\h*(\w+))?      # same as above
(?:,\h*(\w+))?      # same as above
(?:,\h*(\w+))?      # same as above
(?:,\h*(\w+))?      # ... 
(?:,\h*(\w+))?      # ... 
(?:,\h*(\w+))?      # ... 
(?:,\h*(\w+))?      # ... 
(?:,\h*(\w+))?      # ... 

Replacement:

$1              # content of group 1
$2              # content of group 2
\n              # line feed, you can use \r\n for Windows
(?3             # if group 3 exists
    $1              # content of group 1
    $3              # content of group 3
)               # endif
(?4\n$1$4)      # same as above
(?5\n$1$5)      # ...
(?6\n$1$6)      # ...
(?7\n$1$7)      # ...
(?8\n$1$8)      # ...
(?9\n$1$9)      # ...

Screenshot (before):

Screenshot (after):

Rather than using a regex, you can instead extract the first character of the username (using $username[0]) and ltrim() it off the string, and put 1 occurrence of it back in...

$out = $username[0].ltrim($username, $username[0]);