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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 }