feat: add solutions to lc problems: No.2716~2719

* No.2716.Minimize String Length
* No.2717.Semi-Ordered Permutation
* No.2718.Sum of Matrix After Queries
* No.2719.Count of Integers

Author: yanglbme

solution/2500-2599/2559.Count Vowel Strings in Ranges/README.md

@@ -278,7 +278,10 @@ function vowelStrings(words: string[], queries: number[][]): number[] {
     const n = words.length;
     const s: number[] = new Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
-        if (vowels.has(words[i][0]) && vowels.has(words[i][words[i].length - 1])) {
+        if (
+            vowels.has(words[i][0]) &&
+            vowels.has(words[i][words[i].length - 1])
+        ) {
             s[i + 1] = s[i] + 1;
         } else {
             s[i + 1] = s[i];

solution/2500-2599/2559.Count Vowel Strings in Ranges/README_EN.md

@@ -250,7 +250,10 @@ function vowelStrings(words: string[], queries: number[][]): number[] {
     const n = words.length;
     const s: number[] = new Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
-        if (vowels.has(words[i][0]) && vowels.has(words[i][words[i].length - 1])) {
+        if (
+            vowels.has(words[i][0]) &&
+            vowels.has(words[i][words[i].length - 1])
+        ) {
             s[i + 1] = s[i] + 1;
         } else {
             s[i + 1] = s[i];

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

@@ -0,0 +1,128 @@
+# [2716. 最小化字符串长度](https://leetcode.cn/problems/minimize-string-length)
+
+[English Version](/solution/2700-2799/2716.Minimize%20String%20Length/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个下标从 <strong>0</strong> 开始的字符串 <code>s</code> ，重复执行下述操作 <strong>任意</strong> 次：</p>
+
+<ul>
+	<li>在字符串中选出一个下标 <code>i</code> ，并使 <code>c</code> 为字符串下标 <code>i</code> 处的字符。并在 <code>i</code> <strong>左侧</strong>（如果有）和 <strong>右侧</strong>（如果有）各 <strong>删除 </strong>一个距离 <code>i</code> <strong>最近</strong> 的字符 <code>c</code> 。</li>
+</ul>
+
+<p>请你通过执行上述操作任意次，使 <code>s</code> 的长度 <strong>最小化</strong> 。</p>
+
+<p>返回一个表示 <strong>最小化</strong> 字符串的长度的整数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "aaabc"
+<strong>输出：</strong>3
+<strong>解释：</strong>在这个示例中，s 等于 "aaabc" 。我们可以选择位于下标 1 处的字符 'a' 开始。接着删除下标 1 左侧最近的那个 'a'（位于下标 0）以及下标 1 右侧最近的那个 'a'（位于下标 2）。执行操作后，字符串变为 "abc" 。继续对字符串执行任何操作都不会改变其长度。因此，最小化字符串的长度是 3 。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "cbbd"
+<strong>输出：</strong>3
+<strong>解释：</strong>我们可以选择位于下标 1 处的字符 'b' 开始。下标 1 左侧不存在字符 'b' ，但右侧存在一个字符 'b'（位于下标 2），所以会删除位于下标 2 的字符 'b' 。执行操作后，字符串变为 "cbd" 。继续对字符串执行任何操作都不会改变其长度。因此，最小化字符串的长度是 3 。</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "dddaaa"
+<strong>输出：</strong>2
+<strong>解释：</strong>我们可以选择位于下标 1 处的字符 'd' 开始。接着删除下标 1 左侧最近的那个 'd'（位于下标 0）以及下标 1 右侧最近的那个 'd'（位于下标 2）。执行操作后，字符串变为 "daaa" 。继续对新字符串执行操作，可以选择位于下标 2 的字符 'a' 。接着删除下标 2 左侧最近的那个 'a'（位于下标 1）以及下标 2 右侧最近的那个 'a'（位于下标 3）。执行操作后，字符串变为 "da" 。继续对字符串执行任何操作都不会改变其长度。因此，最小化字符串的长度是 2 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 仅由小写英文字母组成</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：哈希表**
+
+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 $C=26$。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def minimizedStringLength(self, s: str) -> int:
+        return len(set(s))
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int minimizedStringLength(String s) {
+        Set<Character> ss = new HashSet<>();
+        for (int i = 0; i < s.length(); ++i) {
+            ss.add(s.charAt(i));
+        }
+        return ss.size();
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minimizedStringLength(string s) {
+        unordered_set<char> ss(s.begin(), s.end());
+        return ss.size();
+    }
+};
+```
+
+### **Go**
+
+```go
+func minimizedStringLength(s string) int {
+	ss := map[rune]struct{}{}
+	for _, c := range s {
+		ss[c] = struct{}{}
+	}
+	return len(ss)
+}
+```
+
+### **TypeScript**
+
+```ts
+function minimizedStringLength(s: string): number {
+    return new Set(s.split('')).size;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

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

@@ -0,0 +1,115 @@
+# [2716. Minimize String Length](https://leetcode.com/problems/minimize-string-length)
+
+[中文文档](/solution/2700-2799/2716.Minimize%20String%20Length/README.md)
+
+## Description
+
+<p>Given a <strong>0-indexed</strong> string <code>s</code>, repeatedly perform the following operation <strong>any</strong> number of times:</p>
+
+<ul>
+	<li>Choose an index <code>i</code> in the string, and let <code>c</code> be the character in position <code>i</code>. <strong>Delete</strong> the <strong>closest occurrence</strong> of <code>c</code> to the <strong>left</strong> of <code>i</code> (if any) and the <strong>closest occurrence</strong> of <code>c</code> to the <strong>right</strong> of <code>i</code> (if any).</li>
+</ul>
+
+<p>Your task is to <strong>minimize</strong> the length of <code>s</code> by performing the above operation any number of times.</p>
+
+<p>Return <em>an integer denoting the length of the <strong>minimized</strong> string.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aaabc&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> In this example, s is &quot;aaabc&quot;. We can start by selecting the character &#39;a&#39; at index 1. We then remove the closest &#39;a&#39; to the left of index 1, which is at index 0, and the closest &#39;a&#39; to the right of index 1, which is at index 2. After this operation, the string becomes &quot;abc&quot;. Any further operation we perform on the string will leave it unchanged. Therefore, the length of the minimized string is 3.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;cbbd&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> For this we can start with character &#39;b&#39; at index 1. There is no occurrence of &#39;b&#39; to the left of index 1, but there is one to the right at index 2, so we delete the &#39;b&#39; at index 2. The string becomes &quot;cbd&quot; and further operations will leave it unchanged. Hence, the minimized length is 3.&nbsp;
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;dddaaa&quot;
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> For this, we can start with the character &#39;d&#39; at index 1. The closest occurrence of a &#39;d&#39; to its left is at index 0, and the closest occurrence of a &#39;d&#39; to its right is at index 2. We delete both index 0 and 2, so the string becomes &quot;daaa&quot;. In the new string, we can select the character &#39;a&#39; at index 2. The closest occurrence of an &#39;a&#39; to its left is at index 1, and the closest occurrence of an &#39;a&#39; to its right is at index 3. We delete both of them, and the string becomes &quot;da&quot;. We cannot minimize this further, so the minimized length is 2.
+</pre>
+
+<div class="notranslate" style="all: initial;">&nbsp;</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> contains only lowercase English letters</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def minimizedStringLength(self, s: str) -> int:
+        return len(set(s))
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int minimizedStringLength(String s) {
+        Set<Character> ss = new HashSet<>();
+        for (int i = 0; i < s.length(); ++i) {
+            ss.add(s.charAt(i));
+        }
+        return ss.size();
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minimizedStringLength(string s) {
+        unordered_set<char> ss(s.begin(), s.end());
+        return ss.size();
+    }
+};
+```
+
+### **Go**
+
+```go
+func minimizedStringLength(s string) int {
+	ss := map[rune]struct{}{}
+	for _, c := range s {
+		ss[c] = struct{}{}
+	}
+	return len(ss)
+}
+```
+
+### **TypeScript**
+
+```ts
+function minimizedStringLength(s: string): number {
+    return new Set(s.split('')).size;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

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

@@ -0,0 +1,7 @@
+class Solution {
+public:
+    int minimizedStringLength(string s) {
+        unordered_set<char> ss(s.begin(), s.end());
+        return ss.size();
+    }
+};

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

@@ -0,0 +1,7 @@
+func minimizedStringLength(s string) int {
+	ss := map[rune]struct{}{}
+	for _, c := range s {
+		ss[c] = struct{}{}
+	}
+	return len(ss)
+}

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

@@ -0,0 +1,9 @@
+class Solution {
+    public int minimizedStringLength(String s) {
+        Set<Character> ss = new HashSet<>();
+        for (int i = 0; i < s.length(); ++i) {
+            ss.add(s.charAt(i));
+        }
+        return ss.size();
+    }
+}

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

@@ -0,0 +1,3 @@
+class Solution:
+    def minimizedStringLength(self, s: str) -> int:
+        return len(set(s))

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

@@ -0,0 +1,3 @@
+function minimizedStringLength(s: string): number {
+    return new Set(s.split('')).size;
+}