How do I approach solving the Student Attendance Record I problem?

Focus on efficiently counting the number of absences and checking for three consecutive late days, both of which can be done in a single pass through the string.

What are the main conditions for the attendance award?

The student needs fewer than two absences and no streak of three consecutive late days.

What is the time complexity of solving this problem?

The time complexity is O(n), where n is the length of the string, because we only need to iterate through the string once.

Can the solution be optimized further?

No, this problem is already optimal with a linear scan approach, as both conditions need to be checked during a single traversal.

Are there edge cases to consider for this problem?

Yes, edge cases include strings with very few characters or strings that contain many consecutive 'L's or 'A's.

#551

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

学生出勤记录 I

给你一个字符串 s 表示一个学生的出勤记录，其中的每个字符用来标记当天的出勤情况（缺勤、迟到、到场）。记录中只含下面三种字符： 'A' ：Absent，缺勤 'L' ：Late，迟到 'P' ：Present，到场如果学生能够同时满足下面两个条件，则可以获得出勤奖励：按总出勤计，学生缺勤…

字符串

题目描述

给你一个字符串 s 表示一个学生的出勤记录，其中的每个字符用来标记当天的出勤情况（缺勤、迟到、到场）。记录中只含下面三种字符：

'A'：Absent，缺勤
'L'：Late，迟到
'P'：Present，到场

如果学生能够同时满足下面两个条件，则可以获得出勤奖励：

按 总出勤 计，学生缺勤（'A'）严格少于两天。
学生不会存在连续 3 天或连续 3 天以上的迟到（'L'）记录。

如果学生可以获得出勤奖励，返回 true ；否则，返回 false 。

示例 1：

输入：s = "PPALLP"
输出：true
解释：学生缺勤次数少于 2 次，且不存在 3 天或以上的连续迟到记录。

示例 2：

输入：s = "PPALLL"
输出：false
解释：学生最后三天连续迟到，所以不满足出勤奖励的条件。

提示：

1 <= s.length <= 1000
s[i] 为 'A'、'L' 或 'P'

lightbulb

解题思路

方法一：字符串遍历

我们可以遍历字符串 $s$ ，记录字符 'A' 和字符串 "LLL" 的出现次数。如果字符 'A' 的出现次数小于 $2$ ，且字符串 "LLL" 没有出现过，则可以将该字符串视作记录合法，返回 true，否则返回 false。

时间复杂度 $O(n)$ ，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$ 。

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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate correctly identifies the need for a linear scan of the string.
question_mark
Candidate efficiently handles the dual check for both 'A' count and 'L' streaks in one pass.
question_mark
Candidate demonstrates awareness of optimal time complexity for this problem.

warning

常见陷阱

外企场景

error
Failing to handle edge cases where the string contains less than three characters.
error
Not returning early if the number of absences exceeds 1 or three consecutive 'L's are detected.
error
Overcomplicating the solution with unnecessary operations, leading to higher time or space complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Check for exactly one absence or at least two consecutive late days.
arrow_right_alt
Determine eligibility based on a maximum of two absences and no 'L' streaks longer than two days.
arrow_right_alt
Extend the logic to check for a specific number of tardies (e.g., exactly one day late allowed).

help

常见问题

外企场景

继续练习

#556 下一个更大元素 III #557 反转字符串中的单词 III #541 反转字符串 II