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feat: add solutions to lc problems: No.0551,0552
* No.0551.Student Attendance Record I * No.0552.Student Attendance Record II
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‎solution/0500-0599/0551.Student Attendance Record I/README.md

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@@ -59,6 +59,26 @@ class Solution {
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}
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```
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### **Go**
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```go
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func checkRecord(s string) bool {
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return strings.Count(s, "A") < 2 && !strings.Contains(s, "LLL")
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool checkRecord(string s) {
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return count(s.begin(), s.end(), 'A') < 2 &&
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s.find("LLL") == string::npos;
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}
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};
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```
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### **...**
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```

‎solution/0500-0599/0551.Student Attendance Record I/README_EN.md

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@@ -69,6 +69,26 @@ class Solution {
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}
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```
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### **Go**
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```go
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func checkRecord(s string) bool {
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return strings.Count(s, "A") < 2 && !strings.Contains(s, "LLL")
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool checkRecord(string s) {
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return count(s.begin(), s.end(), 'A') < 2 &&
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s.find("LLL") == string::npos;
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}
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};
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```
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### **...**
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```

‎solution/0500-0599/0551.Student Attendance Record I/Solution.cpp

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class Solution {
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public:
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bool checkRecord(string s) {
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return count(s.begin(), s.end(), 'A') < 2 &&
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s.find("LLL") == string::npos;
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}
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};

‎solution/0500-0599/0551.Student Attendance Record I/Solution.go

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@@ -0,0 +1,3 @@
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func checkRecord(s string) bool {
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return strings.Count(s, "A") < 2 && !strings.Contains(s, "LLL")
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}

‎solution/0500-0599/0552.Student Attendance Record II/README.md

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@@ -34,22 +34,166 @@
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<!-- 这里可写通用的实现逻辑 -->
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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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- 缺勤：之前不能有任何缺勤记录，即 `j == 0`
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- 迟到：之前最多连续迟到 1 次，即 `k == 0 || k == 1`
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- 到场：无限制
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<!-- tabs:start -->
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### **Python3**
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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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mod = int(1e9 + 7)
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dp = [[[0, 0, 0], [0, 0, 0]] for _ in range(n)]
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# base case
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dp[0][0][0] = dp[0][0][1] = dp[0][1][0] = 1
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for i in range(1, n):
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# A
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dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % mod
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# L
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dp[i][0][1] = dp[i - 1][0][0]
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dp[i][0][2] = dp[i - 1][0][1]
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dp[i][1][1] = dp[i - 1][1][0]
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dp[i][1][2] = dp[i - 1][1][1]
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# P
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dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % mod
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dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % mod
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ans = 0
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for j in range(2):
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for k in range(3):
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ans = (ans + dp[n - 1][j][k]) % mod
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return ans
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```
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### **Java**
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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 static final int MOD = 1000000007;
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public int checkRecord(int n) {
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long[][][] dp = new long[n][2][3];
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// base case
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dp[0][0][0] = 1;
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dp[0][0][1] = 1;
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dp[0][1][0] = 1;
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for (int i = 1; i < n; i++) {
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// A
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dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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// L
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dp[i][0][1] = dp[i - 1][0][0];
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dp[i][0][2] = dp[i - 1][0][1];
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dp[i][1][1] = dp[i - 1][1][0];
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dp[i][1][2] = dp[i - 1][1][1];
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// P
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dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
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}
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long ans = 0;
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for (int j = 0; j < 2; j++) {
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for (int k = 0; k < 3; k++) {
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ans = (ans + dp[n - 1][j][k]) % MOD;
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}
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}
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return (int) ans;
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}
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}
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```
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### **Go**
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```go
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const _mod int = 1e9 + 7
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func checkRecord(n int) int {
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dp := make([][][]int, n)
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for i := 0; i < n; i++ {
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dp[i] = make([][]int, 2)
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for j := 0; j < 2; j++ {
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dp[i][j] = make([]int, 3)
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}
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}
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// base case
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dp[0][0][0] = 1
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dp[0][0][1] = 1
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dp[0][1][0] = 1
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for i := 1; i < n; i++ {
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// A
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dp[i][1][0] = (dp[i-1][0][0] + dp[i-1][0][1] + dp[i-1][0][2]) % _mod
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// L
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dp[i][0][1] = dp[i-1][0][0]
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dp[i][0][2] = dp[i-1][0][1]
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dp[i][1][1] = dp[i-1][1][0]
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dp[i][1][2] = dp[i-1][1][1]
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// P
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dp[i][0][0] = (dp[i-1][0][0] + dp[i-1][0][1] + dp[i-1][0][2]) % _mod
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dp[i][1][0] = (dp[i][1][0] + dp[i-1][1][0] + dp[i-1][1][1] + dp[i-1][1][2]) % _mod
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}
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var ans int
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for j := 0; j < 2; j++ {
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for k := 0; k < 3; k++ {
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ans = (ans + dp[n-1][j][k]) % _mod
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}
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}
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return ans
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}
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```
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### **C++**
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```cpp
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constexpr 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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using ll = long long;
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vector<vector<vector<ll>>> dp(n, vector<vector<ll>>(2, vector<ll>(3)));
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// base case
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dp[0][0][0] = dp[0][0][1] = dp[0][1][0] = 1;
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for (int i = 1; i < n; ++i) {
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// A
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dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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// L
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dp[i][0][1] = dp[i - 1][0][0];
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dp[i][0][2] = dp[i - 1][0][1];
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dp[i][1][1] = dp[i - 1][1][0];
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dp[i][1][2] = dp[i - 1][1][1];
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// P
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dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
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}
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ll ans = 0;
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for (int j = 0; j < 2; ++j) {
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for (int k = 0; k < 3; ++k) {
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ans = (ans + dp[n - 1][j][k]) % MOD;
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}
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}
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return ans;
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}
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};
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```
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### **...**

‎solution/0500-0599/0552.Student Attendance Record II/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -60,13 +60,149 @@ Only &quot;AA&quot; is not eligible because there are 2 absences (there need to
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### **Python3**
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```python
63-
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class Solution:
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def checkRecord(self, n: int) -> int:
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mod = int(1e9 + 7)
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dp = [[[0, 0, 0], [0, 0, 0]] for _ in range(n)]
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# base case
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dp[0][0][0] = dp[0][0][1] = dp[0][1][0] = 1
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for i in range(1, n):
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# A
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dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % mod
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# L
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dp[i][0][1] = dp[i - 1][0][0]
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dp[i][0][2] = dp[i - 1][0][1]
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dp[i][1][1] = dp[i - 1][1][0]
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dp[i][1][2] = dp[i - 1][1][1]
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# P
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dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % mod
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dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % mod
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ans = 0
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for j in range(2):
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for k in range(3):
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ans = (ans + dp[n - 1][j][k]) % mod
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return ans
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```
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### **Java**
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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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public int checkRecord(int n) {
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long[][][] dp = new long[n][2][3];
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// base case
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dp[0][0][0] = 1;
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dp[0][0][1] = 1;
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dp[0][1][0] = 1;
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for (int i = 1; i < n; i++) {
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// A
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dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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// L
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dp[i][0][1] = dp[i - 1][0][0];
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dp[i][0][2] = dp[i - 1][0][1];
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dp[i][1][1] = dp[i - 1][1][0];
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dp[i][1][2] = dp[i - 1][1][1];
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// P
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dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
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}
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long ans = 0;
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for (int j = 0; j < 2; j++) {
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for (int k = 0; k < 3; k++) {
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ans = (ans + dp[n - 1][j][k]) % MOD;
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}
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}
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return (int) ans;
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}
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}
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```
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### **Go**
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```go
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const _mod int = 1e9 + 7
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func checkRecord(n int) int {
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dp := make([][][]int, n)
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for i := 0; i < n; i++ {
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dp[i] = make([][]int, 2)
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for j := 0; j < 2; j++ {
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dp[i][j] = make([]int, 3)
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}
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}
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// base case
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dp[0][0][0] = 1
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dp[0][0][1] = 1
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dp[0][1][0] = 1
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for i := 1; i < n; i++ {
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// A
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dp[i][1][0] = (dp[i-1][0][0] + dp[i-1][0][1] + dp[i-1][0][2]) % _mod
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// L
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dp[i][0][1] = dp[i-1][0][0]
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dp[i][0][2] = dp[i-1][0][1]
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dp[i][1][1] = dp[i-1][1][0]
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dp[i][1][2] = dp[i-1][1][1]
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// P
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dp[i][0][0] = (dp[i-1][0][0] + dp[i-1][0][1] + dp[i-1][0][2]) % _mod
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dp[i][1][0] = (dp[i][1][0] + dp[i-1][1][0] + dp[i-1][1][1] + dp[i-1][1][2]) % _mod
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}
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var ans int
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for j := 0; j < 2; j++ {
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for k := 0; k < 3; k++ {
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ans = (ans + dp[n-1][j][k]) % _mod
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}
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}
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return ans
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}
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```
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### **C++**
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```cpp
173+
constexpr int MOD = 1e9 + 7;
174+
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class Solution {
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public:
177+
int checkRecord(int n) {
178+
using ll = long long;
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vector<vector<vector<ll>>> dp(n, vector<vector<ll>>(2, vector<ll>(3)));
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// base case
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dp[0][0][0] = dp[0][0][1] = dp[0][1][0] = 1;
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for (int i = 1; i < n; ++i) {
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// A
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dp[i][1][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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// L
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dp[i][0][1] = dp[i - 1][0][0];
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dp[i][0][2] = dp[i - 1][0][1];
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dp[i][1][1] = dp[i - 1][1][0];
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dp[i][1][2] = dp[i - 1][1][1];
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// P
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dp[i][0][0] = (dp[i - 1][0][0] + dp[i - 1][0][1] + dp[i - 1][0][2]) % MOD;
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dp[i][1][0] = (dp[i][1][0] + dp[i - 1][1][0] + dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
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}
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ll ans = 0;
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for (int j = 0; j < 2; ++j) {
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for (int k = 0; k < 3; ++k) {
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ans = (ans + dp[n - 1][j][k]) % MOD;
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}
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}
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return ans;
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}
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};
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```
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### **...**

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