Student Attendance Record I

We need to check the two conditions mentioned to determine if the student is eligible for an attendance award:

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


  1. Initialize Counters: We’ll maintain two counters; one to count the total number of absences, and the other to count the current consecutive late days.

  2. Iterate through the Record: We’ll go through the attendance record character by character.

  3. Count Absences: If the character is ‘A’, we’ll increase the absence counter. If the absence counter reaches 2, the student is not eligible, and we can return false immediately.

  4. Count Consecutive Late Days: If the character is ‘L’, we’ll increase the consecutive late counter. If this counter reaches 3, the student is not eligible, and we can return false immediately.

  5. Reset Late Counter for Presence: If the character is ‘P’, we’ll reset the consecutive late counter, as the late streak is broken.

  6. Final Eligibility Check: If we reach the end of the record without violating the conditions, we’ll return true.

Code Implementation

Here’s the code:

class Solution:
    def checkRecord(self, s: str) -> bool:
        absences = 0
        consecutive_late = 0

        for char in s:
            if char == 'A':
                absences += 1
                if absences >= 2:
                    return False
                # Reset late counter
                consecutive_late = 0
            elif char == 'L':
                consecutive_late += 1
                if consecutive_late >= 3:
                    return False
            else: # char is 'P'
                # Reset late counter
                consecutive_late = 0

        return True

Key Takeaways

  • This code efficiently checks the eligibility criteria by iterating through the record once.
  • It uses simple counters to track absences and consecutive late days, making the code easy to understand and maintain.
  • By returning false immediately when a condition is violated, the code avoids unnecessary processing.

This code will determine whether the student’s attendance record qualifies for an award according to the rules provided in the problem statement.

class Solution:
    def checkRecord(self, s: str) -> bool:
    	return (s.count('A') < 2) and ('LLL' not in s)

Problem Classification

This problem falls under String Manipulation and Character Counting.


  1. Input - A string s that represents the attendance record of a student.
  2. Output - A boolean value indicating whether the student is eligible for an attendance award.
  3. Constraints - The student is eligible for an award if they meet the following criteria:
    • They were absent (‘A’) for strictly fewer than 2 days total.
    • They were never late (‘L’) for 3 or more consecutive days.

This is a problem of pattern recognition in a string, tracking the count of specific characters and their consecutive appearances. It could be classified as a Counting problem and a Sliding Window problem (since we’re looking for a specific sequence of characters in the string). It is also a decision problem, as the output is binary (True/False), based on whether certain conditions are met or not.

It requires analyzing a string to identify specific patterns or counts of characters (‘A’ and ‘L’). The domain of the problem is String Manipulation, as it involves traversing and checking for certain patterns within a string. The ‘What’ components are the string itself, and the rules governing whether the student is eligible for an award. It is a decision problem, as we have to decide if the student is eligible or not based on the conditions specified. It could be seen as a Counting problem (count ‘A’s and consecutive ‘L’s) and a Sliding Window problem (look for three consecutive ‘L’s) within the field of algorithms.

Language Agnostic Coding Drills

  1. Dissection of Code Concepts:

a. Defining Functions/Methods: The first distinct concept here is defining a function (or method in the context of a class). It’s about encapsulating a piece of code that performs a specific task into a callable unit. Here, the function checkRecord is defined to check the attendance record.

b. String Handling: The function deals with the string s representing the student’s attendance record.

c. String Counting: One operation performed on the string is count, which counts the number of occurrences of a substring (in this case, the character ‘A’) in the given string.

d. Checking Substring Presence: The function also checks for the presence of a substring (‘LLL’) within the given string.

e. Boolean Operations: The function combines the results of the two string operations with a logical and. It uses the results of these operations to return a single boolean value.

  1. Concepts in Order of Increasing Difficulty:

a. Defining Functions/Methods - This is a basic concept, as it is foundational to most programming tasks.

b. String Handling - Slightly more complex, as it involves understanding how to manipulate and handle strings, which are an essential data type in many languages.

c. String Counting - This concept requires understanding how to count the occurrence of specific characters or substrings in a string, adding a bit more complexity.

d. Checking Substring Presence - Similar in complexity to string counting, this requires understanding how to check for the presence of specific substrings in a string.

e. Boolean Operations - This is a more advanced concept, as it involves understanding logical operators and how to combine boolean values.

  1. Problem-Solving Approach:

The problem is approached by breaking down the given conditions into two separate checks that can be performed on the string:

a. Count the number of ‘A’s in the string and confirm it’s less than 2.

b. Check that the string ‘LLL’ does not occur in the string.

After performing these two checks separately, their results are combined using a logical and operation. This reflects the requirement that both conditions must be met for the student to be eligible for the award.

The two checks align perfectly with the conditions given in the problem, and their combination provides the solution. This approach makes the problem much simpler by allowing us to address each condition separately before combining them. The concepts used to implement these checks, from string handling to boolean operations, all contribute to the final solution.

Targeted Drills in Python

  1. Python Coding Drills for Each Identified Concept:

a. Defining Functions/Methods: This concept can be illustrated through a simple drill where we define a function that squares a number.

def square(n):
    return n * n
print(square(5))  # Outputs: 25

b. String Handling: A basic drill for this could be creating a string and accessing elements from it.

s = "Hello, World!"
print(s[7])  # Outputs: W

c. String Counting: A simple drill can demonstrate how to count the occurrence of a specific character in a string.

s = "Hello, World!"
print(s.count('l'))  # Outputs: 3

d. Checking Substring Presence: A drill could involve checking if a specific word is part of a sentence.

s = "Hello, World!"
print('World' in s)  # Outputs: True

e. Boolean Operations: A drill can show how to combine boolean values using logical operators.

condition1 = True
condition2 = False
print(condition1 and condition2)  # Outputs: False
  1. Problem-Specific Concepts:

The problem-specific concept here is string handling, particularly counting specific characters and checking for the presence of specific substrings. These concepts are vital for our problem as they directly align with the conditions provided in the problem statement: checking the number of ‘A’s and the presence of ‘LLL’ in the string.

  1. Integration of Drills:

To integrate these drills into the final solution:

a. We start by defining a function that takes a string as an argument.

b. We then apply the string counting concept to count the number of ‘A’s in the string and check if it’s less than 2. This gives us one boolean value.

c. Next, we use the concept of checking for substring presence to check if ‘LLL’ is in the string. We negate this value (using not) to align with our requirement (the absence of ‘LLL’). This gives us the second boolean value.

d. Finally, we use a boolean operation (and) to combine these boolean values and return the result.

Each of these steps corresponds to one of our drills, and putting them together in this order forms our solution to the problem.