520_detect_capital.py

class Solution:
    def detectCapitalUse(self, word: str) -> bool:
        if len(word) == 1:
            return True
        upper_chars = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'}
        lower_chars = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'}
        ans = True
        if word[0] in upper_chars:
            if len(word) == 2:
                return True
            else:
                if word[1] in upper_chars:
                    for i in range(2, len(word)):
                        if word[i] not in upper_chars:
                            ans = False
                            break
                else:
                    for i in range(2, len(word)):
                        if word[i] not in lower_chars:
                            ans = False
                            break
        else:
            for i in range(1, len(word)):
                if word[i] not in lower_chars:
                    ans = False
                    break
        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/>.
'''