551 Student Attendance Record I

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

Return True if the studnet is elgible for an attendance award, or False otherwise.

 

constraints :

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

 

2. Algorithms

The two criterias that the studnet is eligible for an attendance award are same as below :

• The total count of absent("A") is fewer than 2.
• The consecutive days of late("L") is fewer than 3.

So when we iterate through the input string 'checkRecord', we need to return False when we encounter error case.

• Return False if count('A') >= 3.
• Return False if s[i : i+3] == 'LLL'.

 

1. Start
2. Declare variable named count initialized with 0.
3. Repeat the steps while loop iterate through integers 0 - len(s) - 1.
   1. If s[i] == 'A', Increment count by 1.
      1. If count > 2, return False
   2. If i < len(s) - 3 and s[i : i+3] == 'LLL', return False
4. Return True when the loop ends.
5. End

 

3. Codes

class Solution:
    def checkRecord(self, s: str) -> bool:
        count = 0
        for i in range(len(s)) : 
            if s[i] == 'A' : 
                count += 1
                if count > 1 : 
                    return False 
            if (i < len(s) - 2) and (s[i:i + 3] == 'LLL') : 
                return False 
        return True

 

4. Conclusion