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Commit ff62ef9

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feat: add solutions to lc problems: No.2712,2716 (doocs#4260)
* No.2712.Minimum Cost to Make All Characters Equal * No.2716.Minimize String Length
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‎solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md‎

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

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

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

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

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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。
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时间复杂度 $O(n),ドル空间复杂度 $O(C)$。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 $C=26$。
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时间复杂度 $O(n),ドル其中 $n$ 是字符串 $\textit{s}$ 的长度。空间复杂度 $O(|\Sigma|),ドル其中 $\Sigma$ 是字符集,这里是小写英文字母,因此 $|\Sigma|=26$。
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<!-- tabs:start -->
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class Solution {
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public:
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int minimizedStringLength(string s) {
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unordered_set<char> ss(s.begin(), s.end());
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return ss.size();
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return unordered_set<char>(s.begin(), s.end()).size();
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}
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};
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```
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}
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```
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#### C#
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```cs
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public class Solution {
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public int MinimizedStringLength(string s) {
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return new HashSet<char>(s).Count;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

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

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### Solution 1: Hash Table
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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.
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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.
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The time complexity is $O(n),ドル 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 $C=26$.
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The time complexity is $O(n),ドル where $n$ is the length of the string $\textit{s}$. The space complexity is $O(|\Sigma|),ドル where $\Sigma$ is the character set. In this case, it's lowercase English letters, so $|\Sigma|=26$.
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<!-- tabs:start -->
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class Solution {
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public:
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int minimizedStringLength(string s) {
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unordered_set<char> ss(s.begin(), s.end());
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return ss.size();
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return unordered_set<char>(s.begin(), s.end()).size();
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}
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};
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```
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}
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```
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#### C#
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```cs
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public class Solution {
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public int MinimizedStringLength(string s) {
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return new HashSet<char>(s).Count;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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class Solution {
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public:
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int minimizedStringLength(string s) {
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unordered_set<char> ss(s.begin(), s.end());
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return ss.size();
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return unordered_set<char>(s.begin(), s.end()).size();
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}
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};
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};
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public class Solution {
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public int MinimizedStringLength(string s) {
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return new HashSet<char>(s).Count;
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}
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}

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