# 552 Student Attendance Record II

An attendance record for a student can be represented as a string 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.

Any 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.

Given an integer `n`, return the number of possible attendance records of length `n` that make a student eligible for an attendance award. The answer may be very large, so return it modulo `109 + 7`.

Example 1:

``````Input: n = 2
Output: 8
Explanation: There are 8 records with length 2 that are eligible for an award:
"PP", "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" is not eligible because there are 2 absences (there need to be fewer than 2).
``````

Example 2:

``````Input: n = 1
Output: 3
``````

Example 3:

``````Input: n = 10101
Output: 183236316
``````

Constraints:

• `1 <= n <= 105`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 `````` ``````class Solution: def checkRecord(self, n: int) -> int: MOD = 10 ** 9 + 7 # dp[n] represents the number of possible # rewardable strings of length n # with L and P as the only characters dp = [1,2,4,7] +  * (2 if n <= 5 else n-3) # dp[i] = 2 * dp[i−1] - dp[i−4] # = (end with P) + (end with L) # - (end with PLLL) for i in range(4, n+1): dp[i] = (2 * dp[i-1]) % MOD + (MOD-dp[i-4]) % MOD # without "A" total = dp[n] # with "A" for i in range(1, n+1): total += (dp[i-1] * dp[n-i]) % MOD return total % MOD``````