Regexp that contains exactly one repetition of 011-CodePudding

I need to write a regular expression that defines a language consisting of all strings of 0s and 1s that contains exactly one substring 011.

This is my approach: 1* 0* (10*)* 011(1)* (0 10)*. I do not know how to fix this, also there is probably more efficient way to solve this..

011 Accept

111000101001011 Accept

110011 Accept

1101011 Accept

110011010 Accept

011011 Accept //it is accepting this one, which is wrong

111111 Reject

101011 Accept

0001011 Accept

CodePudding user response：

Attempt without lookarounds:

^(?>[01]*?011)([01]*?011)? [01]*(?(1)(*FAIL)|)$

try here

Explanation:

^               Match the start of the string
(?>[01]*?011)   Match everything up to and including the first 011, without allowing backtracking
([01]*?011)?    Match everything up to and including the second 011, 
[01]*           Match remaining 1's and 0's
(?(1)(*FAIL)|)  If the second 011 was found, report a failure
$               Match the end of the string

CodePudding user response：

If supported, you can use a negative lookahead asserting not 2 repetitions of 001.

If that is the case, match 011 between optional repetitions of either 0 or 1 using a character class [01]*

^(?![01]*?011[01]*?011)[01]*011[01]*$

See a demo on regex101 for the matches.