find binary gap using regular expression

bingap_regex.py

import re

def bingap(num): # wronggggg
    zeroes = re.findall(r"0+", bin(num)[2:])
    return len(max(zeroes, key=len)) if zeroes else 0

n = 32
print(bin(n))
bingap(n)

This code is actually wrong - -" Bad interpretation of the question.

Binary gap definition is those 0s between two 1s.

This code find the longest running 0s, but they are not necessary being terminated with 1.

ex.
For input 0b100, it will return 2. Which is wrong, as there's no terminating 1 after final 0.
It should return 0, as there is no gap found.)

New corrected version.

import re

def bingap(num):
    zeroes = re.findall(r"0+(?=1)", bin(num)[2:])
    return len(max(zeroes, key=len)) if zeroes else 0

I would add the lookbehind (?<=1) at the beggining, since otherwise you are getting one extra number (the first 1 of each match):

import re

def solution(N):
    zeros = re.findall(r'(?<=1)0+(?=1)', str(bin(N))[2:])
    return len(max(zeros)) if zeros else 0