Checking min no of characters in capturing group in Regex - regex

Question

I have a question in regex
I am dealing with numbers 0 and 1 only
I have 10 digit number grouped into 4 as below
([01]{2})([01]{4})([01]{2})([01]{2})
I need to match all those numbers with min 2 1's in the second group which is ([01]{4}) , no matter how many 0's or 1's other groups are having. I am interested only in the second group
For example, these are the potential matches are
0000110000
0011000000
0001100000
0000110000
I tried using positive look ahead like :
^(\d{2})((?=\d*1{2,}\d*)(\d{4}))(\d{2})(\d{2})
but this is matching even
0000000011
Any help is deeply appreciated

Answer 1

If the two 1s are not necessarily consecutive in Group 2, you can use
^([01]{2})(?=(?:[01]*1){2}[01]{4,6}$)([01]{4})([01]{2})([01]{2})$
See the regex demo
Details:
^ - start of string
([01]{2}) - Group 1: two occurrences of 1 or 0
(?=(?:[01]*1){2}[01]{4,6}$) - immediately to the right of the current location, there must be two occurrences of any zero or more 0 or 1 chars followed with 1 and then there must be four, five or six 1 or 0 chars till the end of string
([01]{4}) - Group 2: four occurrences of 1 or 0
([01]{2}) - Group 3: two occurrences of 1 or 0
([01]{2}) - Group 4: two occurrences of 1 or 0
$ - end of string.

Answer 2

If the ones need to be consecutive (as per your sample data), maybe you can use:
^(?=[01]{2,4}11)[01]{10}$
See the online demo. The idea here is that you would match 2-4 zero's or 1's upto a sequence of two ones. It makes sense if you realise the only combinations that are allowed would have the minimum of two 1's ("11") sequence after exactly 2-4 other digits.
^ - Start line anchor.
(?=[01]{2,4}11) - Open positive lookahead to look for 2-4 characters from our characters class upto "11".
[01]{10} - Match exactly 10 characters from our character class.
$ - End line anchor.
If need be you can change the [01]{10} pieces where you'd use capture groups.
EDIT:
If they don't have to be consecutive, maybe you can work with:
^[01]{2}(?=[01]{8}$)([01]{0,2}1[01]{0,2}1[01]{0,2})[01]{4}$
See the online demo.
Or less verbose:
^(?=[01]{10}$)(..)(.*1.*1.*)(..)(..)$
See the demo

Answer 3

Not a job for regex but for bitwise operators:
(in PHP):
$nums = [
'0000110000',
'0011000000',
'0001100000',
'1000110000',
'0000000110',
'0001000000'
];
foreach ($nums as $num) {
if ( !in_array((bindec($num) >> 4) & 15, [0, 1, 2, 4, 8]) )
echo $num, PHP_EOL;
}
You can probably do that in any language.

Answer 4

If a positive lookahead is supported, you could also assert that group 2 has as least 11 using a positive lookahead.
^([01]{2})(?=[01]{0,2}11)([01]{4})([01]{2})([01]{2})$
^ Start of string
([01]{2}) - Group 1: two occurrences of 1 or 0
(?= Positive lookahead
[01]{0,2}11 Match 0-2 times either 0 or 1 and match 11
) Close lookahead
([01]{4}) - Group 2: four occurrences of 1 or 0
([01]{2}) - Group 3: two occurrences of 1 or 0
([01]{2}) - Group 4: two occurrences of 1 or 0
$ - end of string.
Regex demo
Or you can write out all 3 alternatives matching 11
^([01]{2})(11[01][01]|[01]11[01]|[01][01]11)([01]{2})([01]{2})$
Regex demo