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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

Solution 1

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class Solution:
    def checkRecord(self, n: int) -> int:
        @cache
        def dfs(i, j, k):
            if i >= n:
                return 1
            ans = 0
            if j == 0:
                ans += dfs(i + 1, j + 1, 0)
            if k < 2:
                ans += dfs(i + 1, j, k + 1)
            ans += dfs(i + 1, j, 0)
            return ans % mod

        mod = 10**9 + 7
        ans = dfs(0, 0, 0)
        dfs.cache_clear()
        return ans

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class Solution {
    private final int mod = (int) 1e9 + 7;
    private int n;
    private Integer[][][] f;

    public int checkRecord(int n) {
        this.n = n;
        f = new Integer[n][2][3];
        return dfs(0, 0, 0);
    }

    private int dfs(int i, int j, int k) {
        if (i >= n) {
            return 1;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        int ans = dfs(i + 1, j, 0);
        if (j == 0) {
            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
        }
        if (k < 2) {
            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
        }
        return f[i][j][k] = ans;
    }
}

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class Solution {
public:
    int checkRecord(int n) {
        int f[n][2][3];
        memset(f, -1, sizeof(f));
        const int mod = 1e9 + 7;
        auto dfs = [&](this auto&& dfs, int i, int j, int k) -> int {
            if (i >= n) {
                return 1;
            }
            if (f[i][j][k] != -1) {
                return f[i][j][k];
            }
            int ans = dfs(i + 1, j, 0);
            if (j == 0) {
                ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
            }
            if (k < 2) {
                ans = (ans + dfs(i + 1, j, k + 1)) % mod;
            }
            return f[i][j][k] = ans;
        };
        return dfs(0, 0, 0);
    }
};

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func checkRecord(n int) int {
    f := make([][][]int, n)
    for i := range f {
        f[i] = make([][]int, 2)
        for j := range f[i] {
            f[i][j] = make([]int, 3)
            for k := range f[i][j] {
                f[i][j][k] = -1
            }
        }
    }
    const mod = 1e9 + 7
    var dfs func(i, j, k int) int
    dfs = func(i, j, k int) int {
        if i >= n {
            return 1
        }
        if f[i][j][k] != -1 {
            return f[i][j][k]
        }
        ans := dfs(i+1, j, 0)
        if j == 0 {
            ans = (ans + dfs(i+1, j+1, 0)) % mod
        }
        if k < 2 {
            ans = (ans + dfs(i+1, j, k+1)) % mod
        }
        f[i][j][k] = ans
        return ans
    }
    return dfs(0, 0, 0)
}

Solution 2

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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

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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;
    }
}

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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;
    }
};

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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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