feat: add solutions to lc problem: No.1415

No.1415.The k-th Lexicographical String of All Happy Strings of Length n

Author: yanglbme

solution/1400-1499/1413.Minimum Value to Get Positive Step by Step Sum/README.md

@@ -73,6 +73,13 @@ class Solution:
         return max(1, 1 - t)
 ```
 
+```python
+class Solution:
+    def minStartValue(self, nums: List[int]) -> int:
+        s = list(accumulate(nums))
+        return 1 if min(s) >= 0 else abs(min(s)) + 1
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->

solution/1400-1499/1413.Minimum Value to Get Positive Step by Step Sum/README_EN.md

@@ -65,6 +65,13 @@ class Solution:
         return max(1, 1 - t)
 ```
 
+```python
+class Solution:
+    def minStartValue(self, nums: List[int]) -> int:
+        s = list(accumulate(nums))
+        return 1 if min(s) >= 0 else abs(min(s)) + 1
+```
+
 ### **Java**
 
 ```java

solution/1400-1499/1415.The k-th Lexicographical String of All Happy Strings of Length n/README.md

@@ -69,22 +69,85 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：DFS**
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def getHappyString(self, n: int, k: int) -> str:
+        def dfs(t):
+            if len(t) == n:
+                ans.append(t)
+                return
+            for c in 'abc':
+                if t and t[-1] == c:
+                    continue
+                dfs(t + c)
+
+        ans = []
+        dfs('')
+        return '' if len(ans) < k else ans[k - 1]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    private List<String> ans = new ArrayList<>();
+
+    public String getHappyString(int n, int k) {
+        dfs("", n);
+        return ans.size() < k ? "" : ans.get(k - 1);
+    }
+
+    private void dfs(String t, int n) {
+        if (t.length() == n) {
+            ans.add(t);
+            return;
+        }
+        for (char c : "abc".toCharArray()) {
+            if (t.length() > 0 && t.charAt(t.length() - 1) == c) {
+                continue;
+            }
+            dfs(t + c, n);
+        }
+    }
+}
+```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<string> ans;
+    string getHappyString(int n, int k) {
+        dfs("", n);
+        return ans.size() < k ? "" : ans[k - 1];
+    }
+
+    void dfs(string t, int n) {
+        if (t.size() == n)
+        {
+            ans.push_back(t);
+            return;
+        }
+        for (int c = 'a'; c <= 'c'; ++c)
+        {
+            if (t.size() && t.back() == c) continue;
+            t.push_back(c);
+            dfs(t, n);
+            t.pop_back();
+        }
+    }
+};
 ```
 
 ### **...**

solution/1400-1499/1415.The k-th Lexicographical String of All Happy Strings of Length n/README_EN.md

@@ -57,13 +57,74 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def getHappyString(self, n: int, k: int) -> str:
+        def dfs(t):
+            if len(t) == n:
+                ans.append(t)
+                return
+            for c in 'abc':
+                if t and t[-1] == c:
+                    continue
+                dfs(t + c)
+
+        ans = []
+        dfs('')
+        return '' if len(ans) < k else ans[k - 1]
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    private List<String> ans = new ArrayList<>();
+
+    public String getHappyString(int n, int k) {
+        dfs("", n);
+        return ans.size() < k ? "" : ans.get(k - 1);
+    }
+
+    private void dfs(String t, int n) {
+        if (t.length() == n) {
+            ans.add(t);
+            return;
+        }
+        for (char c : "abc".toCharArray()) {
+            if (t.length() > 0 && t.charAt(t.length() - 1) == c) {
+                continue;
+            }
+            dfs(t + c, n);
+        }
+    }
+}
+```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<string> ans;
+    string getHappyString(int n, int k) {
+        dfs("", n);
+        return ans.size() < k ? "" : ans[k - 1];
+    }
+
+    void dfs(string t, int n) {
+        if (t.size() == n)
+        {
+            ans.push_back(t);
+            return;
+        }
+        for (int c = 'a'; c <= 'c'; ++c)
+        {
+            if (t.size() && t.back() == c) continue;
+            t.push_back(c);
+            dfs(t, n);
+            t.pop_back();
+        }
+    }
+};
 ```
 
 ### **...**

solution/1400-1499/1415.The k-th Lexicographical String of All Happy Strings of Length n/Solution.cpp

@@ -0,0 +1,23 @@
+class Solution {
+public:
+    vector<string> ans;
+    string getHappyString(int n, int k) {
+        dfs("", n);
+        return ans.size() < k ? "" : ans[k - 1];
+    }
+
+    void dfs(string t, int n) {
+        if (t.size() == n)
+        {
+            ans.push_back(t);
+            return;
+        }
+        for (int c = 'a'; c <= 'c'; ++c)
+        {
+            if (t.size() && t.back() == c) continue;
+            t.push_back(c);
+            dfs(t, n);
+            t.pop_back();
+        }
+    }
+};

solution/1400-1499/1415.The k-th Lexicographical String of All Happy Strings of Length n/Solution.java

@@ -0,0 +1,21 @@
+class Solution {
+    private List<String> ans = new ArrayList<>();
+
+    public String getHappyString(int n, int k) {
+        dfs("", n);
+        return ans.size() < k ? "" : ans.get(k - 1);
+    }
+
+    private void dfs(String t, int n) {
+        if (t.length() == n) {
+            ans.add(t);
+            return;
+        }
+        for (char c : "abc".toCharArray()) {
+            if (t.length() > 0 && t.charAt(t.length() - 1) == c) {
+                continue;
+            }
+            dfs(t + c, n);
+        }
+    }
+}

solution/1400-1499/1415.The k-th Lexicographical String of All Happy Strings of Length n/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def getHappyString(self, n: int, k: int) -> str:
+        def dfs(t):
+            if len(t) == n:
+                ans.append(t)
+                return
+            for c in 'abc':
+                if t and t[-1] == c:
+                    continue
+                dfs(t + c)
+
+        ans = []
+        dfs('')
+        return '' if len(ans) < k else ans[k - 1]