7/29 每日一題(如果學生總缺席少於2 且未連續3天以上遲到就返回TRUE)

 You are given a string s representing an attendance record for a student where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:

• 'A': Absent.
• 'L': Late.
• 'P': Present.

The student is eligible for an attendance award if they meet both of the following criteria:

• The student was absent ('A') for strictly fewer than 2 days total.
• The student was never late ('L') for 3 or more consecutive days.

Return true if the student is eligible for an attendance award, or false otherwise.

 

Example 1:

Input: s = "PPALLP"
Output: true
Explanation: The student has fewer than 2 absences and was never late 3 or more consecutive days.

Example 2:

Input: s = "PPALLL"
Output: false
Explanation: The student was late 3 consecutive days in the last 3 days, so is not eligible for the award.

 

Constraints:

• 1 <= s.length <= 1000
• s[i] is either 'A', 'L', or 'P'.

class Solution:

    def checkRecord(self, s: str) -> bool:

        if s.count('A')>=2:

            return False #總共兩天以上缺席

        if re.findall(r'LLL',s):

            return False #連續三天以上遲到

        return True