Regex with one specific number and a word - regex

Question

I would like to match a specific pattern with regex but I am running into catastrophic backtracking. I wonder if there's a way it would be possible to match what I would like and not get an error.
I start with a simple assumption; I want my string to contain only one specific number e.g. 7 and only that specific number:
^\D*7\D*$
Only if I find this pattern do I want to look for another word in the same text such as "Coffee"; I put my condition into a group (^\D*7\D*$) and reference the group in my conditional and the then part will contain "Coffee":
(?(1)Coffee|)
Is there another phrasing that would avoid the the catastrophic backtracking?

Answer 1

You can use a negative lookahead to assert that the word Coffee is at the right.
^(?=.*\bCoffee\b)\D*7\D*$
The pattern matches:
^ Start of string
(?= Positive lookahead, assert that on the right is
.*\bCoffee\b Match Coffee between word boundaries \b to prevent a partial match
) Close lookahead
\D*7\D* Match number 7 between optional non digit characters.
$ End of string
Regex demo
Note that \D also matches a newline. If you don't want to cross newline boundaries, you can use [^\r\n\d] instead.

Answer 2

Left to right checking is more traditional:
^(?=.*Coffee)[^\d7]*7\D*$
See regex proof.
EXPLANATION
--------------------------------------------------------------------------------
^ the beginning of the string
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
Coffee 'Coffee'
--------------------------------------------------------------------------------
) end of look-ahead
--------------------------------------------------------------------------------
[^\d7]* any character except: digits (0-9), '7' (0
or more times (matching the most amount possible))
--------------------------------------------------------------------------------
7 '7'
--------------------------------------------------------------------------------
\D* non-digits (all but 0-9) (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the string
Right to left checking is only possible with engines like latest JavaScript, .NET or PyPi regex in Python:
^[^\d7]*7\D*$(?<=Coffee.*)
See proof.
EXPLANATION
--------------------------------------------------------------------------------
^ the beginning of the string
--------------------------------------------------------------------------------
[^\d7]* any character except: digits (0-9), '7' (0
or more times (matching the most amount
possible))
--------------------------------------------------------------------------------
7 '7'
--------------------------------------------------------------------------------
\D* non-digits (all but 0-9) (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string
--------------------------------------------------------------------------------
(?<= look behind to see if there is:
--------------------------------------------------------------------------------
Coffee 'Coffee'
--------------------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
) end of look-behind