CODEWITHSUNDEEP
regex
Sketchy
Sketchy

Reputation: 31

Match strings that have even number of as and bs and contain substring aabb

I've been trying to get this sorted out for quite some time.

Match strings that have even number of as and bs and contain substring aabb defined over Σ = {a, b}

For the regex flavor, I'm using regexr to build the expression.

I have tried putting the substring between EVEN EVEN but it doesn't work as expected. For example, in even, for:

(aa|bb|(ab|ba)(aa|bb)*(ab|ba))*(aabb)(aa|bb|(ab|ba)(aa|bb)*(ab|ba))*

It doesn't work for strings where there are odd number of a and b before and after the substring. Like for aaabba, it doesn't work even though there are four a characters.

Some words of the language would be:

aaabba, abaabbab, aaaabbbaabbb, ababaabb

Upvotes: 2

Views: 315

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

You may use

^(?=(?:b*ab*a)*b*$)(?=(?:a*ba*b)*a*$)[ab]*aabb[ab]*$

See the JavaScript regex demo

Details

  • ^ - start of string
  • (?=(?:b*ab*a)*b*$) - (even amount of a check) up to the end of string, there can be
    • (?:b*ab*a)* - 0 or more repetitions of two occurrences of 0 or more b chars followed by a and then
    • b* - zero or more b chars
  • (?=(?:a*ba*b)*a*$) - (even amount of b check) up to the end of string, there can be any
    • (?:a*ba*b)* - 0 or more repetitions of two occurrences of 0 or more a chars followed by b and then
    • a* - zero or more a chars
  • [ab]* - 0 or more a or b chars
  • aabb - a required substring
  • [ab]* - 0 or more a or b chars
  • $ - end of string.

Bonus: to match the same with odd amount of a and b chars use

^(?=(?:b*ab*a)*b*ab*$)(?=(?:a*ba*b)*a*ba*$)[ab]*aabb[ab]*$

See the regex demo

Upvotes: 1

Related Questions