feat: add solutions to lc problem: No.2450

No.2450.Number of Distinct Binary Strings After Applying Operations

Author: yanglbme

solution/2400-2499/2450.Number of Distinct Binary Strings After Applying Operations/README.md

@@ -59,22 +59,70 @@ It can be shown that we cannot obtain any other string, so the answer is 2.
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：数学**
+
+假设字符串 $s$ 长度为 $n$，那么长度为 $k$ 的子串共有 $n - k + 1$ 个，每个子串都可以翻转，因此共有 $2^{n - k + 1}$ 种翻转方式。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def countDistinctStrings(self, s: str, k: int) -> int:
+        return pow(2, len(s) - k + 1) % (10**9 + 7)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public static final int MOD = (int) 1e9 + 7;
+
+    public int countDistinctStrings(String s, int k) {
+        int ans = 1;
+        for (int i = 0; i < s.length() - k + 1; ++i) {
+            ans = (ans * 2) % MOD;
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    const int mod = 1e9 + 7;
+
+    int countDistinctStrings(string s, int k) {
+        int ans = 1;
+        for (int i = 0; i < s.size() - k + 1; ++i) {
+            ans = (ans * 2) % mod;
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func countDistinctStrings(s string, k int) int {
+	const mod int = 1e9 + 7
+	ans := 1
+	for i := 0; i < len(s)-k+1; i++ {
+		ans = (ans * 2) % mod
+	}
+	return ans
+}
 ```
 
 ### **TypeScript**

solution/2400-2499/2450.Number of Distinct Binary Strings After Applying Operations/README_EN.md

@@ -60,13 +60,55 @@ It can be shown that we cannot obtain any other string, so the answer is 2.
 ### **Python3**
 
 ```python
-
+class Solution:
+    def countDistinctStrings(self, s: str, k: int) -> int:
+        return pow(2, len(s) - k + 1) % (10**9 + 7)
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public static final int MOD = (int) 1e9 + 7;
+
+    public int countDistinctStrings(String s, int k) {
+        int ans = 1;
+        for (int i = 0; i < s.length() - k + 1; ++i) {
+            ans = (ans * 2) % MOD;
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    const int mod = 1e9 + 7;
+
+    int countDistinctStrings(string s, int k) {
+        int ans = 1;
+        for (int i = 0; i < s.size() - k + 1; ++i) {
+            ans = (ans * 2) % mod;
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func countDistinctStrings(s string, k int) int {
+	const mod int = 1e9 + 7
+	ans := 1
+	for i := 0; i < len(s)-k+1; i++ {
+		ans = (ans * 2) % mod
+	}
+	return ans
+}
 ```
 
 ### **TypeScript**

solution/2400-2499/2450.Number of Distinct Binary Strings After Applying Operations/Solution.cpp

@@ -0,0 +1,12 @@
+class Solution {
+public:
+    const int mod = 1e9 + 7;
+
+    int countDistinctStrings(string s, int k) {
+        int ans = 1;
+        for (int i = 0; i < s.size() - k + 1; ++i) {
+            ans = (ans * 2) % mod;
+        }
+        return ans;
+    }
+};

solution/2400-2499/2450.Number of Distinct Binary Strings After Applying Operations/Solution.go

@@ -0,0 +1,8 @@
+func countDistinctStrings(s string, k int) int {
+	const mod int = 1e9 + 7
+	ans := 1
+	for i := 0; i < len(s)-k+1; i++ {
+		ans = (ans * 2) % mod
+	}
+	return ans
+}

solution/2400-2499/2450.Number of Distinct Binary Strings After Applying Operations/Solution.java

@@ -0,0 +1,11 @@
+class Solution {
+    public static final int MOD = (int) 1e9 + 7;
+
+    public int countDistinctStrings(String s, int k) {
+        int ans = 1;
+        for (int i = 0; i < s.length() - k + 1; ++i) {
+            ans = (ans * 2) % MOD;
+        }
+        return ans;
+    }
+}

solution/2400-2499/2450.Number of Distinct Binary Strings After Applying Operations/Solution.py

@@ -0,0 +1,3 @@
+class Solution:
+    def countDistinctStrings(self, s: str, k: int) -> int:
+        return pow(2, len(s) - k + 1) % (10**9 + 7)