Description
552. Student Attendance Record II (Hard)
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
Solution
We denote dp2[i][0] as the number of cases of Late on the i + 1th day, dp2[i][1] as the number of cases of Late on the i + 1th day, when there are only two cases: Late and Present.
dp2[i][1] = dp2[i - 1][0] + dp2[i - 1][1];dp2[i][0] = dp2[i - 1][1] + dp2[i - 1][0] - dp2[n - 3][1];
We denote dp[i][0] as the number of cases of Late on the i + 1th day, dp[i][1] as the number of cases of Late on the i + 1th day, dp[i][2] as the number of cases of Absent on the i + 1th day.
dp[i][1] = dp[i - 1][0] + dp[i - 1][1] + dp[i - 1][2];dp[i][2] = dp2[i - 1][0] + dp2[i - 1][1];dp[i][0] = dp[i - 1][1] + dp[i - 1][2] + dp[i - 1][0] - (dp[i - 3][1] + dp[i - 3][2]);
Code
class Solution {
public:
int checkRecord(int n) {
if (n == 1)
return 3;
if (n == 2)
return 8;
if (n == 3)
return 19;
vector<vector<long long>> dp3(n + 1, vector<long long>(3, 0));
vector<vector<long long>> dp2(n, vector<long long>(2, 0));
int mod = 1000000007;
dp2[0][0] = dp2[0][1] = 1;
dp2[1][0] = dp2[1][1] = 2;
dp2[2][0] = 3;
dp2[2][1] = 4;
for (int i = 3; i < n; i++) {
dp2[i][1] = (dp2[i - 1][0] % mod + dp2[i - 1][1] % mod) % mod;
dp2[i][0] = (dp2[i - 1][1] % mod + (dp2[i - 1][0] - dp2[i - 3][1] + mod) % mod) % mod;
}
dp3[0][1] = dp3[0][0] = dp3[0][2] = 1;
dp3[1][0] = 3, dp3[1][1] = 3, dp3[1][2] = 2;
dp3[2][0] = 3 + 2 + 3 - 1, dp3[2][1] = 8, dp3[2][2] = 4;
for (int i = 1; i < n; i++) {
dp3[i][2] = (dp2[i - 1][0] % mod + dp2[i - 1][1] % mod) % mod;
}
for (int i = 3; i < n; i++) {
dp3[i][0] = (dp3[i - 1][1] % mod + dp3[i - 1][2] % mod + (dp3[i - 1][0] - (dp3[i - 3][1] + dp3[i - 3][2]) + mod) % mod) % mod;
dp3[i][1] = (dp3[i - 1][1] % mod + dp3[i - 1][0] % mod + dp3[i - 1][2] % mod) % mod;
}
return (dp3[n - 1][0] + dp3[n - 1][1] + dp3[n - 1][2]) % mod;
}
};