LeetCode 0551. Student Attendance Record I Solution in Java, C++, Python & More | Explanation + Code

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.

 

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'.

Solutions

Solution 1

PythonJavaC++GoTypeScript

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

class Solution { public boolean checkRecord(String s) { return s.indexOf("A") == s.lastIndexOf("A") && !s.contains("LLL"); } }(code-box)

class Solution { public: bool checkRecord(string s) { return count(s.begin(), s.end(), 'A') < 2 && s.find("LLL") == string::npos; } };(code-box)

func checkRecord(s string) bool { return strings.Count(s, "A") < 2 && !strings.Contains(s, "LLL") }(code-box)

function checkRecord(s: string): boolean { return s.indexOf('A') === s.lastIndexOf('A') && s.indexOf('LLL') === -1; }(code-box)