feat: add solutions to lc problems: No.2712,2716 (#4260)

* No.2712.Minimum Cost to Make All Characters Equal
* No.2716.Minimize String Length

Author: yanglbme

solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md

@@ -156,6 +156,24 @@ function minimumCost(s: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn minimum_cost(s: String) -> i64 {
+        let mut ans = 0;
+        let n = s.len();
+        let s = s.as_bytes();
+        for i in 1..n {
+            if s[i] != s[i - 1] {
+                ans += i.min(n - i);
+            }
+        }
+        ans as i64
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README_EN.md

@@ -155,6 +155,24 @@ function minimumCost(s: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn minimum_cost(s: String) -> i64 {
+        let mut ans = 0;
+        let n = s.len();
+        let s = s.as_bytes();
+        for i in 1..n {
+            if s[i] != s[i - 1] {
+                ans += i.min(n - i);
+            }
+        }
+        ans as i64
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/Solution.rs

@@ -0,0 +1,13 @@
+impl Solution {
+    pub fn minimum_cost(s: String) -> i64 {
+        let mut ans = 0;
+        let n = s.len();
+        let s = s.as_bytes();
+        for i in 1..n {
+            if s[i] != s[i - 1] {
+                ans += i.min(n - i);
+            }
+        }
+        ans as i64
+    }
+}

solution/2700-2799/2716.Minimize String Length/README.md

@@ -72,7 +72,7 @@ tags:
 
 题目实际上可以转化为求字符串中不同字符的个数，因此，我们只需要统计字符串中不同字符的个数即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串的长度；而 $C$ 是字符集的大小，本题中字符集为小写英文字母，因此 $C=26$。
+时间复杂度 $O(n)$，其中 $n$ 是字符串 $\textit{s}$ 的长度。空间复杂度 $O(|\Sigma|)$，其中 $\Sigma$ 是字符集，这里是小写英文字母，因此 $|\Sigma|=26$。
 
 <!-- tabs:start -->
 
@@ -104,8 +104,7 @@ class Solution {
 class Solution {
 public:
     int minimizedStringLength(string s) {
-        unordered_set<char> ss(s.begin(), s.end());
-        return ss.size();
+        return unordered_set<char>(s.begin(), s.end()).size();
     }
 };
 ```
@@ -143,6 +142,16 @@ impl Solution {
 }
 ```
 
+#### C#
+
+```cs
+public class Solution {
+    public int MinimizedStringLength(string s) {
+        return new HashSet<char>(s).Count;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2700-2799/2716.Minimize String Length/README_EN.md

@@ -100,9 +100,9 @@ tags:
 
 ### Solution 1: Hash Table
 
-The problem can actually be transformed into finding the number of different characters in the string. Therefore, we only need to count the number of different characters in the string.
+The problem can actually be transformed into finding the number of distinct characters in the string. Therefore, we only need to count the number of distinct characters in the string.
 
-The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, the character set is lowercase English letters, so $C=26$.
+The time complexity is $O(n)$, where $n$ is the length of the string $\textit{s}$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this case, it's lowercase English letters, so $|\Sigma|=26$.
 
 <!-- tabs:start -->
 
@@ -134,8 +134,7 @@ class Solution {
 class Solution {
 public:
     int minimizedStringLength(string s) {
-        unordered_set<char> ss(s.begin(), s.end());
-        return ss.size();
+        return unordered_set<char>(s.begin(), s.end()).size();
     }
 };
 ```
@@ -173,6 +172,16 @@ impl Solution {
 }
 ```
 
+#### C#
+
+```cs
+public class Solution {
+    public int MinimizedStringLength(string s) {
+        return new HashSet<char>(s).Count;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2700-2799/2716.Minimize String Length/Solution.cpp

@@ -1,7 +1,6 @@
 class Solution {
 public:
     int minimizedStringLength(string s) {
-        unordered_set<char> ss(s.begin(), s.end());
-        return ss.size();
+        return unordered_set<char>(s.begin(), s.end()).size();
     }
-};
+};

solution/2700-2799/2716.Minimize String Length/Solution.cs

@@ -0,0 +1,5 @@
+public class Solution {
+    public int MinimizedStringLength(string s) {
+        return new HashSet<char>(s).Count;
+    }
+}