使用正则表达式匹配具有奇数个 0 和偶数个 1 的二进制字符串

Question

[Redacted Info]

Some valid inputs

• 101
• 10001
• 00011
• 11000
• 0

Some invalid Inputs

• 11
• 00
• 10101
• 101101111

EDIT: For the people suggesting that ReGeX is not the way to go are absolutely correct, but for this question I need to use regex. Also, My definition of simpler is reducing the number of characters in the regex.(The minimum is about 22 characters long)

Answer 1

If you're determined to do this with regex, then for certain definitions of "simplify", this may fit the bill.

(?=^0*((10*){2})*?$)(?=^1*(01*)((01*){2})*?$)^.*$

(?=                )                                assert that
   ^              $                                 between the start and end of the string
    0*                                              (consume leading zeros)
      (        )*?                                  there appears as many times as necessary
            {2}                                     two instances of 
       (10*)                                        a 1 followed by any number of 0s
                    (?=^1*     ((01*){2})*?$)       perform the same check as before
                          (01*)                     but require an extra 0 at the start

This relies on the {2} quantifier to demand a multiple of 2 for the numbers in question, and instead of validating the string all at once, performs 2 checks on the string: the first looks for an even number of 1s, and the second looks for an even number of 0s, plus an extra 0.

Demo

Answer 2

Using Counter from collections is another way to come at it. It makes it easy to look for odd and even sequences as well as ensuring that the string has only 1s and 0s.

from collections import Counter

def is_valid_binary_string(test_string):
    c = Counter(test_string)
    
    # Not valid if anything other than 1s and 0s
    if set(c.keys()) - {"0", "1"}:
        return False
    
    # Valid only if even number of 1s and odd number of 0s
    return c["1"] % 2 == 0 and c["0"] % 2 == 1

Answer 3

You do not need regex for this. Its super simple to achieve this using string.count() and checking if the response is odd or even.

testcases = ['101', '10001', '00011', '11000', '0', '11', '00', '10101', '101101111']
for string in testcases:
    print(string, 'Valid' if string.count('0') % 2 == 1 and string.count('1') % 2 == 0 else 'Invalid')

Outputs

101 Valid
10001 Valid
00011 Valid
11000 Valid
0 Valid
11 Invalid
00 Invalid
10101 Invalid
101101111 Invalid