Python: Find longest binary gap in binary representation of an integer number

Your implementation converts the integer to a base two string then visits each character in the string. Instead, you could just visit each bit in the integer using << and &. Doing so will avoid visiting each bit twice (first to convert it to a string, then to check if if it's a "1" or not in the resulting string). It will also avoid allocating memory for the string and then for each substring you inspect.

You can inspect each bit of the integer by visiting 1 << 0, 1 << 1, ..., 1 << (x.bit_length).

For example:

def max_gap(x):
    max_gap_length = 0
    current_gap_length = 0
    for i in range(x.bit_length()):
        if x & (1 << i):
            # Set, any gap is over.
            if current_gap_length > max_gap_length:
                max_gap_length = current_gap_length
            current_gap_length = 0
         else:
            # Not set, the gap widens.
            current_gap_length += 1
    # Gap might end at the end.
    if current_gap_length > max_gap_length:
        max_gap_length = current_gap_length
    return max_gap_length

def max_gap(N):
    xs = bin(N)[2:].strip('0').split('1')
    return max([len(x) for x in xs])

Explanation:

1. Both leading and trailing zeros are redundant with binary gap finding as they are not bounded by two 1's (left and right respectively)
2. So step 1 striping zeros left and right
3. Then splitting by 1's yields all sequences of 0'z
4. Solution: The maximum length of 0's sub-strings

I do realize that brevity does not mean readability nor efficiency.

However, ability to spell out solution in spoken language and implement it in Python in no time constitutes efficient use of my time.

For binary gap: hey, lets convert int into binary, strip trailing zeros, split at '1' to list, then find longest element in list and get this element lenght.

def binary_gap(N):
    return len(max(format(N, 'b').strip('0').split('1')))