Regex for a pattern XXYYZZ

Question

I want to validate a string which should follow the pattern XXYYZZ where X, Y, Z can be any letter a-z, A-Z or 0-9.

Example of valid strings:

RRFFKK
BB7733
WWDDMM
5599AA

Not valid:

555677
AABBCD

For now I am splitting the string using the regex (?<=(.))(?!\\1) and iterating over the resulting array and checking if each substring has a length of 2.

String str = "AABBEE";
boolean isValid = checkPattern(str);

public static boolean checkPattern(String str) {
   String splited = str.split("(?<=(.))(?!\\1)");
   for (String s : splited) {
       if (s.length() != 2) {
         return false;
       }
   }
   return true;
}

I would like to replace my way of checking with String#matches and get rid of the loop, but can't come up with a valid regex. Can some one help what to put in someRegex in the below snippet?

public static boolean checkPattern(String str) {
    return str.matches(someRegex);
}

Answer 1

You can use

s.matches("(\\p{Alnum})\\1(?!\\1)(\\p{Alnum})\\2(?!\\1|\\2)(\\p{Alnum})\\3")

See the regex demo.

Details

• \A - start of string (it is implicit in String#matches) - the start of string
• (\p{Alnum})\1 - an alphanumeric char (captured into Group 1) and an identical char right after
• (?!\1) - the next char cannot be the same as in Group 1
• (\p{Alnum})\2 - an alphanumeric char (captured into Group 2) and an identical char right after
• (?!\1|\2) - the next char cannot be the same as in Group 1 and 2
• (\p{Alnum})\3 - an alphanumeric char (captured into Group 3) and an identical char right after
• \z - (implicit in String#matches) - end of string.

RegexPlanet test results:

Answer 2

Since you know a valid pattern will always be six characters long with three pairs of equal characters which are different from each other, a short series of explicit conditions may be simpler than a regex:

public static boolean checkPattern(String str) {
   return str.length() == 6 &&
          str.charAt(0) == str.chatAt(1) &&
          str.charAt(2) == str.chatAt(3) &&
          str.charAt(4) == str.chatAt(5) &&
          str.charAt(0) != str.charAt(2) &&
          str.charAt(0) != str.charAt(4) &&
          str.charAt(2) != str.charAt(4);
}

Answer 3

Would the following work for you?

^(([A-Za-z\d])\2(?!.*\2)){3}$

See the online demo

---

• ^ - Start string anchor.
• (- Open 1st capture group.
  • ( - Open 2nd capture group.
    • [A-Za-z\d] - Any alphanumeric character.
    • ) - Close 2nd capture group.
  • \2 - Match exactly what was just captured.
  • (?!.*\2) - Negative lookahead to make sure the same character is not used elsewhere.
  • ) - Close 1st capture group.
• {3} - Repeat the above three times.
• $ - End string anchor.

Answer 4

Well, here's another solution that uses regex and streams in combination.

• It breaks up the pattern into groups of two characters.
• keeps the distinct groups.
• and returns true if the count is 3.

String[] data = { "AABBBB", "AABBCC", "AAAAAA","AABBAA", "ABC", "AAABCC",
        "RRABBCCC" };
String pat = "(?:\\G(.)\\1)+";
Pattern pattern = Pattern.compile(pat);

for (String str : data) {
    Matcher m = pattern.matcher(str);
    boolean isValid = m.results().map(MatchResult::group).distinct().count() == 3;
    System.out.printf("%8s  ->  %s%n",
            str, isValid ? "Valid" : "Not Valid");
}

Prints

  AABBBB  ->  Not Valid
  AABBCC  ->  Valid
  AAAAAA  ->  Not Valid
  AABBAA  ->  Not Valid
     ABC  ->  Not Valid
  AAABCC  ->  Not Valid
RRABBCCC  ->  Not Valid