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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
    }
    

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