Regular expression for exact one character occurrence at any place of the string

Question

I am writing a Java code that finds a way out from any maze and I need a class that checks generated mazes. Only ., #, S, X characters are allowed.

1. . and # - one and more occurrences at any place of the string are allowed

2. S and X - one and not more than one occurrence at any place of the string is allowed.

^[#//.]+$ - regex for the first condition, But I cannot implement the second one.

The maze input looks like this:

.......S..#.
.....###....
..X.........

. - empty space, # - wall, S - start, X - exit

Answer 1

You can use negative lookahead groups, written like (?!...), to accomplish this, like so:

^(?!.*S.*S)(?!.*X.*X)[SX.#]+$

Demo

This accepts any set of characters from your set (S, X, ., #) from the start of the string using the ^ and [SX.#]+. But it rejects any string containing 2 Ss ((?!.*S.*S)) or 2 Xs ((?!.*X.*X)).

Note that this actually checks both of your conditions. You don't really need 2 regexes here. Based on your example maze, though, it looks like your input can span multiple lines. In that case, you need to add \n inside the final character class.