696_count_binary_substrings.py

class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        if len(s) <= 1:
            return 0
        ans = last = 0
        cur = 1
        for i in range(1, len(s)):
            if s[i] == s[i-1]:
                cur += 1
            else:
                last = cur
                cur = 1
            if last >= cur:
                ans += 1
        return ans

'''
    This is my personal record of solving Leetcode Problems. 
    If you have any questions, please discuss them in [Issues](https://github.com/mengxinayan/leetcode/issues).
    Copyright (C) 2020-2022  mengxinayan

    This program is free software: you can redistribute it and/or modify
    it under the terms of the GNU General Public License as published by
    the Free Software Foundation, either version 3 of the License, or
    (at your option) any later version.

    This program is distributed in the hope that it will be useful,
    but WITHOUT ANY WARRANTY; without even the implied warranty of
    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
    GNU General Public License for more details.

    You should have received a copy of the GNU General Public License
    along with this program.  If not, see <http://www.gnu.org/licenses/>.
'''