README.md

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn check_record(n: i32) -> i32 {
        let n = n as usize;
        let mut dp = vec![[[0_i64; 3]; 2]; n + 1];
        let mut ret = 0;
        dp[0][0][0] = 1;

        for i in 0..n {
            dp[i + 1][1][0] = dp[i][0][0] + dp[i][0][1] + dp[i][0][2];
            for j in 0..2 {
                dp[i + 1][j][0] += dp[i][j][0] + dp[i][j][1] + dp[i][j][2];
                dp[i + 1][j][0] %= 1_000_000_007;
                for k in 0..2 {
                    dp[i + 1][j][k + 1] += dp[i][j][k];
                    dp[i + 1][j][k + 1] %= 1_000_000_007;
                }
            }
        }

        for j in 0..2 {
            for k in 0..3 {
                ret = (ret + dp[n][j][k] as i32) % 1_000_000_007;
            }
        }

        ret
    }
}