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
CodePudding user response：
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
CodePudding user response：
If you don't mind repeating both "forms", an alternation is probably the most straightforward solution:
\b(?:([1-9])0\10\10|0([1-9])0\20\2)\b