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feat: add solutions to lc problem: No.0467
No.0467.Unique Substrings in Wraparound String
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‎solution/0400-0499/0467.Unique Substrings in Wraparound String/README.md

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@@ -53,9 +53,13 @@
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<!-- 这里可写通用的实现逻辑 -->
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题目导读，为什么示例 2 只有两个字串？
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**方法一：动态规划**
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> 成为字串的一个标准，需要是连续的，`a` 与 `c` 之间少了一个 `b`，所以不能算一个子字符串
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不妨设 `dp[α]` 表示 p 中以字符 α 结尾且在 s 中的子串的最大长度，将 dp 求和可以得到最终结果。
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时间复杂度 O(n)，其中 n 表示字符串 p 的长度。
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> 成为子串的一个标准，需要是连续的，`a` 与 `c` 之间少了一个 `b`，所以不能算一个子字符串。
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<!-- tabs:start -->
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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 findSubstringInWraproundString(self, p: str) -> int:
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dp = [0] * 26
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k = 0
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for i, c in enumerate(p):
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if i and (ord(c) - ord(p[i - 1])) % 26 == 1:
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k += 1
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else:
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k = 1
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idx = ord(c) - ord('a')
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dp[idx] = max(dp[idx], k)
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return sum(dp)
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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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public int findSubstringInWraproundString(String p) {
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int[] dp = new int[26];
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int k = 0;
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for (int i = 0; i < p.length(); ++i) {
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char c = p.charAt(i);
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if (i > 0 && (c - p.charAt(i - 1) + 26) % 26 == 1) {
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++k;
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} else {
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k = 1;
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}
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dp[c - 'a'] = Math.max(dp[c - 'a'], k);
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}
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int ans = 0;
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for (int v : dp) {
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ans += v;
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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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### **TypeScript**
@@ -120,6 +154,58 @@ impl Solution {
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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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int findSubstringInWraproundString(string p) {
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vector<int> dp(26);
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int k = 0;
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for (int i = 0; i < p.size(); ++i)
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{
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char c = p[i];
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if (i && (c - p[i - 1] + 26) % 26 == 1) ++k;
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else k = 1;
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dp[c - 'a'] = max(dp[c - 'a'], k);
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}
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int ans = 0;
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for (int& v : dp) ans += v;
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return 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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func findSubstringInWraproundString(p string) int {
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dp := make([]int, 26)
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k := 0
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for i := range p {
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c := p[i]
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if i > 0 && (c-p[i-1]+26)%26 == 1 {
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k++
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} else {
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k = 1
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}
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dp[c-'a'] = max(dp[c-'a'], k)
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}
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ans := 0
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for _, v := range dp {
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ans += v
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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```

‎solution/0400-0499/0467.Unique Substrings in Wraparound String/README_EN.md

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@@ -52,13 +52,43 @@ Explanation: Only the substring &quot;a&quot; of p is in s.
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### **Python3**
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```python
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class Solution:
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def findSubstringInWraproundString(self, p: str) -> int:
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dp = [0] * 26
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k = 0
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for i, c in enumerate(p):
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if i and (ord(c) - ord(p[i - 1])) % 26 == 1:
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k += 1
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else:
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k = 1
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idx = ord(c) - ord('a')
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dp[idx] = max(dp[idx], k)
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return sum(dp)
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```
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### **Java**
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```java
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class Solution {
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public int findSubstringInWraproundString(String p) {
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int[] dp = new int[26];
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int k = 0;
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for (int i = 0; i < p.length(); ++i) {
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char c = p.charAt(i);
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if (i > 0 && (c - p.charAt(i - 1) + 26) % 26 == 1) {
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++k;
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} else {
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k = 1;
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}
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dp[c - 'a'] = Math.max(dp[c - 'a'], k);
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}
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int ans = 0;
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for (int v : dp) {
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ans += v;
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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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### **TypeScript**
@@ -106,6 +136,58 @@ impl Solution {
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}
107137
```
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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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int findSubstringInWraproundString(string p) {
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vector<int> dp(26);
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int k = 0;
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for (int i = 0; i < p.size(); ++i)
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{
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char c = p[i];
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if (i && (c - p[i - 1] + 26) % 26 == 1) ++k;
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else k = 1;
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dp[c - 'a'] = max(dp[c - 'a'], k);
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}
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int ans = 0;
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for (int& v : dp) ans += v;
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return 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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func findSubstringInWraproundString(p string) int {
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dp := make([]int, 26)
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k := 0
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for i := range p {
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c := p[i]
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if i > 0 && (c-p[i-1]+26)%26 == 1 {
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k++
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} else {
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k = 1
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}
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dp[c-'a'] = max(dp[c-'a'], k)
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}
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ans := 0
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for _, v := range dp {
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ans += v
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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```

‎solution/0400-0499/0467.Unique Substrings in Wraparound String/Solution.cpp

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@@ -0,0 +1,17 @@
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class Solution {
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public:
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int findSubstringInWraproundString(string p) {
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vector<int> dp(26);
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int k = 0;
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for (int i = 0; i < p.size(); ++i)
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{
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char c = p[i];
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if (i && (c - p[i - 1] + 26) % 26 == 1) ++k;
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else k = 1;
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dp[c - 'a'] = max(dp[c - 'a'], k);
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}
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int ans = 0;
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for (int& v : dp) ans += v;
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return ans;
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}
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};

‎solution/0400-0499/0467.Unique Substrings in Wraparound String/Solution.go

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func findSubstringInWraproundString(p string) int {
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dp := make([]int, 26)
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k := 0
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for i := range p {
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c := p[i]
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if i > 0 && (c-p[i-1]+26)%26 == 1 {
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k++
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} else {
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k = 1
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}
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dp[c-'a'] = max(dp[c-'a'], k)
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}
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ans := 0
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for _, v := range dp {
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ans += v
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/0400-0499/0467.Unique Substrings in Wraparound String/Solution.java

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class Solution {
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public int findSubstringInWraproundString(String p) {
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int[] dp = new int[26];
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int k = 0;
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for (int i = 0; i < p.length(); ++i) {
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char c = p.charAt(i);
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if (i > 0 && (c - p.charAt(i - 1) + 26) % 26 == 1) {
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++k;
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} else {
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k = 1;
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}
12+
dp[c - 'a'] = Math.max(dp[c - 'a'], k);
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}
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int ans = 0;
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for (int v : dp) {
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ans += v;
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}
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return ans;
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}
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}

‎solution/0400-0499/0467.Unique Substrings in Wraparound String/Solution.py

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class Solution:
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def findSubstringInWraproundString(self, p: str) -> int:
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dp = [0] * 26
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k = 0
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for i, c in enumerate(p):
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if i and (ord(c) - ord(p[i - 1])) % 26 == 1:
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k += 1
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else:
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k = 1
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idx = ord(c) - ord('a')
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dp[idx] = max(dp[idx], k)
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return sum(dp)

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