# 552. Student Attendance Record II

## 1. Questions

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.

## 2. Examples

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


## 3. Constraints

• 1 <= n <= 105

## 5. Solutions

class Solution {
public int checkRecord(int n) {
final int MOD = 1000000007;
int[][][] dp = new int[n + 1][2][3]; // 长度，A 的数量，结尾连续 L 的数量
dp[0][0][0] = 1;
for (int i = 1; i <= n; i++) {
// 以 P 结尾的数量
for (int j = 0; j <= 1; j++) {
for (int k = 0; k <= 2; k++) {
dp[i][j][0] = (dp[i][j][0] + dp[i - 1][j][k]) % MOD;
}
}
// 以 A 结尾的数量
for (int k = 0; k <= 2; k++) {
dp[i][1][0] = (dp[i][1][0] + dp[i - 1][0][k]) % MOD;
}
// 以 L 结尾的数量
for (int j = 0; j <= 1; j++) {
for (int k = 1; k <= 2; k++) {
dp[i][j][k] = (dp[i][j][k] + dp[i - 1][j][k - 1]) % MOD;
}
}
}
int sum = 0;
for (int j = 0; j <= 1; j++) {
for (int k = 0; k <= 2; k++) {
sum = (sum + dp[n][j][k]) % MOD;
}
}
return sum;
}
}