Replace whitespace in part of a string

Question

I have cases which contain urls in the case description, and the urls contain whitespace. I need to replace the whitespace with %20 in order to make the links clickable from the case. Obviously the rest of the case description before/after the URLs should remain untouched.

I'm trying to test how to do this in dev console just using an input string for the initial investigation, so far I have:

String inputString = 'John Doe \n Attachments: Download link: \\\\link\\TRA001AAA\\021840136\\some receipt 2021-02-16.pdf \n Claim submission reference number: CS6207713.';
pattern myPattern = pattern.compile('TRA001AAA(.*)'); 
matcher myMatcher = myPattern.matcher(inputString);
System.debug('myMatcher.find() = '+ myMatcher.find());
System.debug('myMatcher.group() = '+ myMatcher.group());
String regExp = '[\\s]';
String noSpaces = myMatcher.group().replaceAll(regExp, '%20');
System.debug('noSpaces = ' + noSpaces);

The debug output is:

DEBUG|myMatcher.find() = true
15:08:46:007 USER_DEBUG [5]|DEBUG|myMatcher.group() = TRA001AAA\021840136\some receipt 2021-02-16.pdf
15:08:46:007 USER_DEBUG [8]|DEBUG|noSpaces = TRA001AAA\021840136\some%20receipt%202021-02-16.pdf%20

I have two problems now

1. Theres a trailing space in my match, causing the extra %20 at the end
2. How do I get this back into the original text? So the desired final string would be: 'John Doe \n Attachments: Download link: \\\\link\\TRA001AAA\\021840136\\some%20 receipt%202021-02-16.pdf \n Claim submission reference number: CS12345.