Enclose code in anonymous brackets saves memory?

Question

I'm a beginner and for some reason, I like to make effort to save memory. For now I think it's fun.

So in this problem, I have a regex that matches kebab-case words and later turns them into camel-case words, and I want to know if that is some occurrence in a given string. I tried Matcher.matches() but I found out that it only works if the entire string is matched, so I only could think of compiling the regex and use the Matcher.find() method to put a boolean value inside a variable, but I wanted to enclose everything in brackets to save memory.

This is my solution:

String regex = "(?:([\\p{IsAlphabetic}]*)?(-[\\p{IsAlphabetic}]+))+";

boolean hasSubSequence;
{
    Matcher m = Pattern.compile(regex).matcher(identifier);
    hasSubSequence = m.find();
}

if (hasSubSequence) {
    Matcher kebabCaseMatches = Pattern.compile(regex).matcher(identifier);
    while (kebabCaseMatches.find()) {
        String currentOccurence = kebabCaseMatches.group();
        
        while (currentOccurence.contains("-")) {
            currentOccurence = currentOccurence.replaceFirst("-[\\p{IsAlphabetic}]", Character.toString(Character.toUpperCase(currentOccurence.charAt(currentOccurence.indexOf("-") + 1))));
        }
        
        // "identifier" is the function's argument
        identifier = identifier.replaceFirst(regex, currentOccurence);
    }
}

Does this really save memory?

Answer 1

You could compile the pattern once where you declared regex (static final?): would save speed and many instances. Brackets are voodoo.