CODEWITHSUNDEEP
javaregexstring
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Reputation: 840

Java - Delete each 4th occurrence of a character in a row

I'm searching for a way to delete each 4th occurrence of a character (a-zA-Z) in a row.

For example, if I have the following string:
helloooo I am veeeeeeeeery busy right nowww because I am working veeeeeery hard

I want delete all 4th, 5th, 6th, ... characters in a row. But, in the word hard, a 4th r occurs, which I do NOT want to delete, because it is not the 4th r in a row / it is surrounded with other characters. The result should be: hellooo I am veeery busy right nowww because I am working veeery hard

I have already searched for a way to do this, and I could have found a way to replace/delete the 4th occurrence of a character, but I could not find a way to replace/delete the 4th occurrence of a character in a row.

Thanks in advance.

Upvotes: 2

Views: 108

Answers (3)

Sebastian Lenartowicz
Sebastian Lenartowicz

Reputation: 4874

The regex you want is ((.)\2{2})\2*. Not quite sure what that is in Java-ese, but what it does is match any single character and then 2 additional instances of that character, followed by any number of additional instances. Then replace it with the contents of the first capture group (\1) and you're good to go.

Upvotes: 2

Vladimir Petrakovich
Vladimir Petrakovich

Reputation: 4286

The function may be written like this:

public static String transform(String input) {
    if (input.isEmpty()) {
        return input;
    } else {
        final StringBuilder sb = new StringBuilder();
        char lastChar = '\0';
        int duplicates = 0;
        for (int i = 0; i < input.length(); i++) {
            final char curChar = input.charAt(i);
            if (curChar == lastChar) {
                duplicates++;
                if (duplicates < 3) {
                    sb.append(curChar);
                }
            } else {
                sb.append(curChar);
                lastChar = curChar;
                duplicates = 0;
            }
        }
        return sb.toString();
    }
}

I think it's faster than regex.

Upvotes: 3

anubhava
anubhava

Reputation: 785846

In Java you can use this replacement based on back-references:

str = str.replaceAll("(([a-zA-Z])\\2\\2)\\2+", "$1");

Code Demo

RegEx Demo

Upvotes: 2

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