Is Student Attendance Record I easy or hard?

Student Attendance Record I is categorized as an Easy problem on LeetCode. The challenge focuses on careful string iteration and constraint checking rather than advanced data structures or algorithms.

Student Attendance Record I Python/Java solution

In Python or Java, the typical solution iterates through the string and maintains two counters: one for 'A' occurrences and one for consecutive 'L' characters. If the absence count reaches two or the late streak reaches three, the function returns false. The same O(n) logic applies across Python, Java, C++, and JavaScript.

How to solve Student Attendance Record I in O(n)?

Iterate through the string once while maintaining two variables: an absence counter and a consecutive late counter. Increment the absence counter for 'A'. Increment the late streak for 'L' and reset it for other characters. If absences reach 2 or the late streak reaches 3, return false; otherwise return true after the scan.

What is the best approach for Student Attendance Record I?

The best approach is a single-pass scan of the string while tracking two counters: total absences and the current consecutive late streak. If absences reach 2 or the late streak reaches 3, the record is invalid. This method runs in O(n) time and O(1) space and is the solution most interviewers expect.

Is Student Attendance Record I asked at Google/Amazon/Meta?

This problem represents a common string validation pattern often used in interviews at companies like Amazon and other large tech firms. While the exact question may vary, the underlying concept of scanning a string and enforcing constraints with counters appears frequently in coding interviews.

What data structure is used in Student Attendance Record I?

No complex data structure is required. The solution works directly on the input string with simple integer counters to track absences and consecutive late days. This keeps memory usage constant and the implementation straightforward.

What is the time complexity of Student Attendance Record I?

The optimal solution runs in O(n) time where n is the length of the attendance string. Each character is processed once while maintaining counters for absences and consecutive late days. Space complexity is O(1) because only a few integer variables are required.

Description Solution Video 4:49

551. Student Attendance Record I

Easy50.1% acceptance3 companies

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'.

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551. Student Attendance Record I

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