552 Student Attendance Record II

šŸ”€ Pick One

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
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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] + [0] * (2 if n <= 5 else n-3)
        # dp[i] = 2 * dp[iāˆ’1] - dp[iāˆ’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