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feat: add solutions to lc problem: No.0552
No.0552.Student Attendance Record II
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‎solution/0500-0599/0552.Student Attendance Record II/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：记忆化搜索**
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我们设计一个函数 $dfs(i, j, k)$，表示从第 $i$ 个出勤记录开始，当前缺勤次数为 $j$，目前最后连续迟到次数为 $k$ 时，可获得出勤奖励的情况数量。那么答案就是 $dfs(0, 0, 0)$。
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函数 $dfs(i, j, k)$ 的执行过程如下：
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- 如果 $i \ge n$，说明已经遍历完所有出勤记录，返回 $1$；
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- 如果 $j = 0$，说明当前缺勤次数为 $0$，那么可以选择缺勤，即 $dfs(i + 1, j + 1, 0)$；
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- 如果 $k \lt 2$，说明当前连续迟到次数小于 $2$，那么可以选择迟到，即 $dfs(i + 1, j, k + 1)$；
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- 无论如何，都可以选择到场，即 $dfs(i + 1, j, 0)$。
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我们将上述三种情况的结果相加，即为 $dfs(i, j, k)$ 的结果。
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为了避免重复计算，我们可以使用记忆化搜索。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为出勤记录的长度。
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**方法二：动态规划**
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动态规划，定义 `dp[i][j][k]` 表示前 `i` 天，缺勤 `j` 次，连续迟到 `k` 次时，可获得出勤奖励的情况数量
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状态转移需要对第 `i` 天的出勤情况分别讨论：
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def checkRecord(self, n: int) -> int:
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@cache
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def dfs(i, j, k):
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if i >= n:
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return 1
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ans = 0
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if j == 0:
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ans += dfs(i + 1, j + 1, 0)
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if k < 2:
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ans += dfs(i + 1, j, k + 1)
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ans += dfs(i + 1, j, 0)
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return ans % mod
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mod = 10**9 + 7
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ans = dfs(0, 0, 0)
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dfs.cache_clear()
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return ans
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```
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```python
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class Solution:
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def checkRecord(self, n: int) -> int:
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private final int mod = (int) 1e9 + 7;
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private int n;
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private Integer[][][] f;
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public int checkRecord(int n) {
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this.n = n;
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f = new Integer[n][2][3];
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return dfs(0, 0, 0);
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}
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private int dfs(int i, int j, int k) {
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if (i >= n) {
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return 1;
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}
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if (f[i][j][k] != null) {
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return f[i][j][k];
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}
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int ans = dfs(i + 1, j, 0);
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if (j == 0) {
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ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
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}
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if (k < 2) {
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ans = (ans + dfs(i + 1, j, k + 1)) % mod;
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}
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return f[i][j][k] = ans;
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}
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}
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```
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```java
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class Solution {
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private static final int MOD = 1000000007;
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### **Go**
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```go
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func checkRecord(n int) int {
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f := make([][][]int, n)
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for i := range f {
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f[i] = make([][]int, 2)
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for j := range f[i] {
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f[i][j] = make([]int, 3)
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for k := range f[i][j] {
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f[i][j][k] = -1
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}
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}
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}
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const mod = 1e9 + 7
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var dfs func(i, j, k int) int
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dfs = func(i, j, k int) int {
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if i >= n {
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return 1
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}
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if f[i][j][k] != -1 {
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return f[i][j][k]
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}
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ans := dfs(i+1, j, 0)
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if j == 0 {
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ans = (ans + dfs(i+1, j+1, 0)) % mod
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}
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if k < 2 {
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ans = (ans + dfs(i+1, j, k+1)) % mod
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}
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f[i][j][k] = ans
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return ans
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}
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return dfs(0, 0, 0)
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}
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```
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```go
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const _mod int = 1e9 + 7
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### **C++**
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```cpp
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int f[100010][2][3];
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const int mod = 1e9 + 7;
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class Solution {
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public:
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int checkRecord(int n) {
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this->n = n;
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memset(f, -1, sizeof(f));
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return dfs(0, 0, 0);
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}
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int dfs(int i, int j, int k) {
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if (i >= n) {
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return 1;
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}
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if (f[i][j][k] != -1) {
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return f[i][j][k];
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}
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int ans = dfs(i + 1, j, 0);
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if (j == 0) {
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ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
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}
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if (k < 2) {
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ans = (ans + dfs(i + 1, j, k + 1)) % mod;
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}
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return f[i][j][k] = ans;
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}
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private:
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int n;
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};
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```
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```cpp
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constexpr int MOD = 1e9 + 7;
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‎solution/0500-0599/0552.Student Attendance Record II/README_EN.md

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### **Python3**
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```python
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class Solution:
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def checkRecord(self, n: int) -> int:
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@cache
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def dfs(i, j, k):
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if i >= n:
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return 1
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ans = 0
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if j == 0:
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ans += dfs(i + 1, j + 1, 0)
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if k < 2:
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ans += dfs(i + 1, j, k + 1)
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ans += dfs(i + 1, j, 0)
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return ans % mod
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mod = 10**9 + 7
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ans = dfs(0, 0, 0)
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dfs.cache_clear()
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return ans
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```
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```python
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class Solution:
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def checkRecord(self, n: int) -> int:
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### **Java**
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```java
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class Solution {
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private final int mod = (int) 1e9 + 7;
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private int n;
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private Integer[][][] f;
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public int checkRecord(int n) {
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this.n = n;
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f = new Integer[n][2][3];
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return dfs(0, 0, 0);
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}
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private int dfs(int i, int j, int k) {
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if (i >= n) {
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return 1;
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}
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if (f[i][j][k] != null) {
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return f[i][j][k];
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}
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int ans = dfs(i + 1, j, 0);
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if (j == 0) {
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ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
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}
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if (k < 2) {
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ans = (ans + dfs(i + 1, j, k + 1)) % mod;
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}
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return f[i][j][k] = ans;
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}
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}
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```
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```java
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class Solution {
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private static final int MOD = 1000000007;
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### **Go**
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```go
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func checkRecord(n int) int {
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f := make([][][]int, n)
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for i := range f {
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f[i] = make([][]int, 2)
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for j := range f[i] {
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f[i][j] = make([]int, 3)
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for k := range f[i][j] {
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f[i][j][k] = -1
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}
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}
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}
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const mod = 1e9 + 7
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var dfs func(i, j, k int) int
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dfs = func(i, j, k int) int {
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if i >= n {
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return 1
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}
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if f[i][j][k] != -1 {
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return f[i][j][k]
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}
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ans := dfs(i+1, j, 0)
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if j == 0 {
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ans = (ans + dfs(i+1, j+1, 0)) % mod
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}
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if k < 2 {
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ans = (ans + dfs(i+1, j, k+1)) % mod
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}
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f[i][j][k] = ans
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return ans
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}
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return dfs(0, 0, 0)
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}
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```
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```go
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const _mod int = 1e9 + 7
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### **C++**
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```cpp
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int f[100010][2][3];
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const int mod = 1e9 + 7;
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class Solution {
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public:
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int checkRecord(int n) {
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this->n = n;
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memset(f, -1, sizeof(f));
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return dfs(0, 0, 0);
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}
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int dfs(int i, int j, int k) {
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if (i >= n) {
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return 1;
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}
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if (f[i][j][k] != -1) {
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return f[i][j][k];
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}
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int ans = dfs(i + 1, j, 0);
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if (j == 0) {
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ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
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}
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if (k < 2) {
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ans = (ans + dfs(i + 1, j, k + 1)) % mod;
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}
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return f[i][j][k] = ans;
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}
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private:
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int n;
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};
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```
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```cpp
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constexpr int MOD = 1e9 + 7;
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