[Leetcode a daily question] 925. Long press to enter (string comparison)

Leetcode’s daily question
link: 925. Long press to enter the
problem-solving idea: use two pointers name_index and typed_index to traverse two strings respectively, divided into the following three situations:

- 当前两个字符相同并且下一个字符也相同时，同时向后移动一个字符
- 当前两个字符相同并且下一个字符不相同时，typed_index 向后移动一个字符
- 其他情况则为False

Finally, judge whether the size of name_index and typed_index is consistent with the length of the string. There is a little trick, because you want to judge the next character, you can add $ at the end of the character to prevent subscript overflow and judge the last character.
answer:

class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        if len(name) == 0:
            return True
        if len(typed) == 0:
            return False

        name += "$"
        typed += "$"
        len_name = len(name) - 1
        len_typed = len(typed) - 1

        name_index, typed_index = 0, 0
        while name_index < len_name and typed_index < len_typed:
            if (name[name_index] == typed[typed_index]) and (name[name_index + 1] == typed[typed_index + 1]):
                typed_index += 1
                name_index += 1
            elif (name[name_index] == typed[typed_index]) and (name[name_index + 1] != typed[typed_index + 1]):
                typed_index += 1
            else:
                break
            
        # print(name_index, typed_index)
        return name_index == len_name and typed_index == len_typed