Student Attendance Record I

Problem 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.

Key Insights

• The attendance record consists of three characters: 'A', 'L', and 'P'.
• We need to count the occurrences of 'A' and check for 'L' in consecutive sequences.
• The constraints provide a manageable string length (up to 1000), allowing for a linear scan approach.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string s.
Space Complexity: O(1), as we only use a constant amount of space for counters and flags.

Solution

To solve the problem, we will perform a single pass through the string while maintaining:

• A counter for the number of absences ('A').
• A counter to track consecutive late days ('L').

The algorithm works as follows:

1. Initialize a counter for absences and a counter for consecutive lates.
2. Iterate through each character in the string:
   • If the character is 'A', increment the absence counter.
   • If the character is 'L', increment the consecutive late counter.
   • If the character is 'P', reset the consecutive late counter to zero.
3. If at any point the absence counter exceeds 1 or the consecutive late counter reaches 3, return false.
4. If the loop completes without these conditions being met, return true.

Code Solutions