Date: 2026-06-06
Time: 19:18
This file solves LeetCode 551 — Student Attendance Record I. It determines whether a student qualifies for an attendance award based on two disqualifying conditions in their attendance string:
1. 2 or more absences ('A') total
2. 3 or more consecutive lates ('L')
The string consists only of characters 'A' (absent), 'L' (late), and 'P' (present).
Solution.checkRecord(self, s: str) -> boolSingle-pass classifier over the attendance string. Returns True if the student is eligible (passes both constraints), False otherwise.
Contract: Input is a string of {'A', 'L', 'P'} characters. Output is a boolean. No mutation, no side effects.
Early-exit accumulator: The method uses two counters (absences, consecutive_lates) and returns False the moment either threshold is reached. This avoids scanning the rest of the string once a disqualifying condition is found.
State reset on transition: consecutive_lates is reset to 0 whenever a non-'L' character is encountered (both 'A' and 'P' branches). This is correct because the problem asks for *consecutive* lates — any interruption breaks the streak.
Note that absences is monotonically increasing (never reset), while consecutive_lates is reset on every non-'L' character. This reflects the difference between the two rules: absences are global, consecutive lates are local.
Imports: None. Pure standard Python.
Imported by: The test_solution.py in the same directory. The massive "Imported By" list in the prompt is misleading — those are unrelated test files in sibling problem directories that happen to share the same module naming convention. They import their own solution.py, not this one.
1. Initialize absences = 0, consecutive_lates = 0
2. For each character c in s:
'A': increment absences, return False if >= 2, reset consecutive_lates'L': increment consecutive_lates, return False if >= 3'P' (the else): reset consecutive_lates only3. If the loop completes without returning False, return True
Complexity: O(n) time, O(1) space — single pass with two integer counters.
absences equals the count of 'A' characters seen so far.consecutive_lates equals the length of the current unbroken run of 'L' characters ending at the current position (or 0 if the current character isn't 'L').False at the earliest possible moment — it never processes a character after a disqualifying condition is detected.None. The function trusts its input is a valid attendance string. Characters outside {'A', 'L', 'P'} silently fall into the else branch, behaving like 'P' (resetting the late counter, not incrementing absences). This is fine for LeetCode's constrained input guarantees.
student-attendance-record-i/test_solution.py — Validates edge cases like strings with exactly 2 A's, exactly 3 consecutive L's, and boundary combinationsstudent-attendance-record-i/plan.md — Documents the problem decomposition and approach selection before implementationstudent-attendance-record-ii — LeetCode 552 is the hard DP variant of this problem (count valid strings of length n), a significant complexity jumpdelete-characters-to-make-fancy-string/solution.py:Solution.makeFancyString — Uses a similar consecutive-character tracking pattern (no 3 consecutive identical chars)single-pass-with-early-exit — A recurring pattern across this repo for problems with disqualifying conditionsearly-exit-on-disqualify — checkRecord returns False immediately upon finding 2 absences or 3 consecutive lates, never scanning further than necessaryconsecutive-late-reset-on-any-non-L — Both 'A' and 'P' reset consecutive_lates to 0, meaning absences break a late streak (which matches the problem specification)unknown-chars-treated-as-present — Any character not 'A' or 'L' falls into the else branch and behaves identically to 'P'o1-time-o1-space — The solution is O(n) time and O(1) space with exactly two integer counters