The numbers should be 6 digit and of the form 0x0x0x or x0x0x0, where x can be any digit from 1 to 9. Ex – 202020, 030303, 808080, etc.
I have this regex that matches numbers with alternative 0 and 1s, cannot make it work for the above use case
b(?!d*(d)1)[10]+b
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Answer
The easiest way is probably to capture the repeating part.
Then check if the back reference to group 1 is repeated.
b(0[1-9]|[1-9]0)1{2}b