# 552. Student Attendance Record II

## Description

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

• class Solution {
private static final int MOD = 1000000007;

public int checkRecord(int n) {
long[][][] dp = new long[n][2][3];

// base case
dp[0][0][0] = 1;
dp[0][0][1] = 1;
dp[0][1][0] = 1;

for (int i = 1; i < n; i++) {
// A
dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
// L
dp[i][0][1] = dp[i - 1][0][0];
dp[i][0][2] = dp[i - 1][0][1];
dp[i][1][1] = dp[i - 1][1][0];
dp[i][1][2] = dp[i - 1][1][1];
// P
dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
}

long ans = 0;
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 3; k++) {
ans = (ans + dp[n - 1][j][k]) % MOD;
}
}
return (int) ans;
}
}


• constexpr int MOD = 1e9 + 7;

class Solution {
public:
int checkRecord(int n) {
using ll = long long;
vector<vector<vector<ll>>> dp(n, vector<vector<ll>>(2, vector<ll>(3)));

// base case
dp[0][0][0] = dp[0][0][1] = dp[0][1][0] = 1;

for (int i = 1; i < n; ++i) {
// A
dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
// L
dp[i][0][1] = dp[i - 1][0][0];
dp[i][0][2] = dp[i - 1][0][1];
dp[i][1][1] = dp[i - 1][1][0];
dp[i][1][2] = dp[i - 1][1][1];
// P
dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
}

ll ans = 0;
for (int j = 0; j < 2; ++j) {
for (int k = 0; k < 3; ++k) {
ans = (ans + dp[n - 1][j][k]) % MOD;
}
}
return ans;
}
};


• class Solution:
def checkRecord(self, n: int) -> int:
mod = int(1e9 + 7)
dp = [[[0, 0, 0], [0, 0, 0]] for _ in range(n)]

# base case
dp[0][0][0] = dp[0][0][1] = dp[0][1][0] = 1

for i in range(1, n):
# A
dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % mod
# L
dp[i][0][1] = dp[i - 1][0][0]
dp[i][0][2] = dp[i - 1][0][1]
dp[i][1][1] = dp[i - 1][1][0]
dp[i][1][2] = dp[i - 1][1][1]
# P
dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % mod
dp[i][1][0] = (
dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]
) % mod

ans = 0
for j in range(2):
for k in range(3):
ans = (ans + dp[n - 1][j][k]) % mod
return ans


• const _mod int = 1e9 + 7

func checkRecord(n int) int {
dp := make([][][]int, n)
for i := 0; i < n; i++ {
dp[i] = make([][]int, 2)
for j := 0; j < 2; j++ {
dp[i][j] = make([]int, 3)
}
}

// base case
dp[0][0][0] = 1
dp[0][0][1] = 1
dp[0][1][0] = 1

for i := 1; i < n; i++ {
// A
dp[i][1][0] = (dp[i-1][0][0] + dp[i-1][0][1] + dp[i-1][0][2]) % _mod
// L
dp[i][0][1] = dp[i-1][0][0]
dp[i][0][2] = dp[i-1][0][1]
dp[i][1][1] = dp[i-1][1][0]
dp[i][1][2] = dp[i-1][1][1]
// P
dp[i][0][0] = (dp[i-1][0][0] + dp[i-1][0][1] + dp[i-1][0][2]) % _mod
dp[i][1][0] = (dp[i][1][0] + dp[i-1][1][0] + dp[i-1][1][1] + dp[i-1][1][2]) % _mod
}

var ans int
for j := 0; j < 2; j++ {
for k := 0; k < 3; k++ {
ans = (ans + dp[n-1][j][k]) % _mod
}
}
return ans
}