551. Student Attendance Record I #

Problem #

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'.

Problem Summary #

You are given a string s representing a student’s attendance record, where each character marks the attendance status for that day (absent, late, or present). The record contains only the following three characters:

• ‘A’: Absent
• ‘L’: Late
• ‘P’: Present

If the student can satisfy both of the following conditions at the same time, they can receive an attendance award:

• In terms of total attendance, the student was absent (‘A’) for strictly fewer than two days.
• The student does not have a record of being late (‘L’) for 3 or more consecutive days.

If the student can receive an attendance award, return true; otherwise, return false.

Solution Approach #

• Traverse the string s to find the total number of ‘A’s and the maximum number of consecutive ‘L’s.
• Check whether the number of ‘A’s is less than 2 and the maximum consecutive number of ‘L’s is less than 3.

Code #

package leetcode

func checkRecord(s string) bool {
	numsA, maxL, numsL := 0, 0, 0
	for _, v := range s {
		if v == 'L' {
			numsL++
		} else {
			if numsL > maxL {
				maxL = numsL
			}
			numsL = 0
			if v == 'A' {
				numsA++
			}
		}
	}
	if numsL > maxL {
		maxL = numsL
	}
	return numsA < 2 && maxL < 3
}