re findall() with overlapping == true

Here’s a question that asks for 0+ between 1

https://leetcode.com/problems/binary-gap/

If you use re.findall(), you’ll get an error 

binary = 11111111010011110100
result = re.findall(r"1(0+)1", binary)

the result is ['0', '0']

If you use your own function, you’ll get it right

def findall_overlapped(r, s):
    res = []                     # Resulting list
    reg = r'^{}$'.format(r)      # Regex must match full string
    for q in range(len(s)):      # Iterate over all chars in a string
        for w in range(q,len(s)):  # Iterate over the rest of the chars to the right
            cur = s[q:w+1]         # Currently tested slice
            if re.match(reg, cur): # If there is a full slice match
                res.append(cur)    # Append it to the resulting list
    return res

rex = r'1(0+)1'
print(findall_overlapped(rex, binary))

the result is ['101', '1001', '101']

If you use c++ with symbols of ?=, you can also get it right

        string input_seq = "11111111010011110100";
        std::regex re("1(?=(0+)1)"); // <-- working
        //std::regex re("1(0+)1"); // <-- not work
        std::sregex_iterator next(input_seq.begin(), input_seq.end(), re);
        std::sregex_iterator end;
        while (next != end)
        {
            std::smatch match = *next;
            std::cout << match.str(1) << "\t"
                      << "\t" << match.position() << "\t"
                      << "\n"; // <-- SEE HERE
            next++;
        }

the result is ['0', '00', '00']

The explanation

According to the python document, is:
(?=...) Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion.

For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.

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