LeetCode 0552. Student Attendance Record II Solution in Java, C++, Python & Go | Explanation + Code

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

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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); } };(code-box)

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) }(code-box)

Solution 2

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)