Set minimum and maximum characters in a regular expression

Yes

Just like + means one or more you can use {3,30} to match between 3 and 30

For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters

From the documentation of the Pattern class

X{n,m}    X, at least n but not more than m times

In your case, matching 3-30 letters followed by spaces could be accomplished with:

([a-zA-Z]\s){3,30}

If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)

([a-zA-Z]\s){2,29}[a-zA-Z]

If you'd like whitespaces to count as characters you need to divide that number by 2 to get

([a-zA-Z]\s){1,14}[a-zA-Z]

You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet

If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) at the beginning of the RegExp and removing the other size limitations

All that said, in all honestly I'd probably just test the String's .length property. It's more readable.

This is what you are looking for

^[a-zA-Z](\s?[a-zA-Z]){2,29}$

^ is the start of string

$ is the end of string

(\s?[a-zA-Z]){2,29} would match (\s?[a-zA-Z]) 2 to 29 times..

Actually Benjamin's answer will lead to the complete solution to the OP's question. Using lookaheads it is possible to restrict the total number of characters AND restrict the match to a set combination of letters and (optional) single spaces.

The regex that solves the entire problem would become

(?=^.{3,30}$)^([A-Za-z][\s]?)+$

This will match AAA, A A and also fail to match AA A since there are two consecutive spaces. I tested this at http://regexpal.com/ and it does the trick.