Define a regex, which matches one digit twice and all others once - ruby-on-rails

Question

As part of a larger regex I would like to match the following restrictions:
The string has 11 digits
All digits are numbers
Within the first 10 digits one number [0-9] (and one only!) must be listed twice
This means the following should match:
12345678914
12235879600
Whereas these should not:
12345678903 -> none of the numbers at digits 1 to 10 appears twice
14427823482 -> one number appears more than twice
72349121762 -> two numbers appear twice
I have tried to use a lookahead, but all I'm managing is that the regex counts a certain digit, i.e.:
(?!.*0\1{2})
That does not do what I need. Is my query even possible with regex?

Answer 1

You can use this kind of pattern:
\A(?=\d{11}\z)(?:(\d)(?!\d*\1\d))*(\d)(?=\d*\2\d)(?:(\d)(?!\d*\3\d))+\d\z
online demo
pattern details:
the idea is to describe string as a duplicate digit surrounded by non duplicate digits.
Finding a duplicate digit is easy with a capture group, a lookahead assertion and a backreference:(\d)(?=\d*\1)
You can use the same pattern to ensure that a digit has no duplicate, but this time with a negative lookahead: (\d)(?!\d*\1)
To not take in account the last digit (digit n°11) in the search of duplicates, you only need to add a digit after the backreference. (\d)(?=\d*\1\d) (in this way you ensure there is at least one digit between the backreference and the end of the string.)
Note that in the present context, what is called a duplicate digit is a digit that is not followed immediatly or later with the same digit. (i.e. in 1234567891 the first 1 is a duplicate digit, but the last 1 is no more a duplicate digit because it is not followed by an other 1)
\A # begining of the string
(?=\d{11}\z) # check the string length (if not needed, remove it)
(?:(\d)(?!\d*\1\d))* # zero or more non duplicate digits
(\d)(?=\d*\2\d) # one duplicate digit
(?:(\d)(?!\d*\3\d))+ # one or more non duplicate digits
\d # the ignored last digit
\z # end of the string
an other way
This time you check the duplicates at the begining of the pattern with lookaheads. One lookahead to ensure there is one duplicate digit, one negative lookahead to ensure there are not two duplicate digits:
\A(?=\d*(\d)(?=\d*\1\d))(?!\d*(\d)(?=\d*\2\d)\d*(\d)(?=\d*\3\d))\d{11}\z
pattern details:
\A
(?= # check if there is one duplicate digit
\d*(\d)(?=\d*\1\d)
)
(?! # check if there are not two duplicate digits
\d*(\d)(?=\d*\2\d) # the first
\d*(\d)(?=\d*\3\d) # the second
)
\d{11}
\z
Note: However it seems that the first way is more efficient.
The code way
You can easily check if your string fit the requirements with array methods:
> mydigs = "12345678913"
=> "12345678913"
> puts (mydigs.split(//).take 10).uniq.size == 9
true
=> nil