551. Student Attendance Record I

Description

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'.

Submitted Code

class Solution(object):
    def checkRecord(self, s):
        """
        :type s: str
        :rtype: bool
        """
        a_count = 0
        l_count = 0

        for i in range(len(s)):
            if s[i] == "P":
                l_count = 0         # 연속 지각일 초기화
            if s[i] == "A":
                l_count = 0         # 연속 지각일 초기화
                a_count += 1        # 결석횟수 + 1
                if a_count == 2:        # 총 지각횟수 2회가 되는 순간 개근상 x
                    return False
            elif s[i] == "L":
                l_count += 1        # 연속 지각일 + 1
                if l_count == 3:        # 연속 지각일 3일이 되는 순간 개근상 x
                    return False
        
        return True

Runtime: 0 ms | Beats 100.00%
Memory: 12.48 MB | Beats 53.08%

“P”, “A”인 경우에는 l_count를 다시 0으로 초기화하는 방식으로 연속 지각일을 체크했다.

Other Solutions

1st

class Solution:
    def checkRecord(self, s: str) -> bool:
        return (s.count('A')<2) and len(s.split('LLL'))==1

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)

반복문을 사용하지 않아도 count(), split() 함수로 풀 수 있다. 문자열 내에 “LLL”이 있다면 split했을 때 최소 2개의 문자열로 나뉘어지기 때문에 False가 된다.