feat: add new lc problems

Author: yanglbme

solution/2200-2299/2268.Minimum Number of Keypresses/README.md

@@ -0,0 +1,94 @@
+# [2268. Minimum Number of Keypresses](https://leetcode.cn/problems/minimum-number-of-keypresses)
+
+[English Version](/solution/2200-2299/2268.Minimum%20Number%20of%20Keypresses/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>You have a keypad with <code>9</code> buttons, numbered from <code>1</code> to <code>9</code>, each mapped to lowercase English letters. You can choose which characters each button is matched to as long as:</p>
+
+<ul>
+	<li>All 26 lowercase English letters are mapped to.</li>
+	<li>Each character is mapped to by <strong>exactly</strong> <code>1</code> button.</li>
+	<li>Each button maps to <strong>at most</strong> <code>3</code> characters.</li>
+</ul>
+
+<p>To type the first character matched to a button, you press the button once. To type the second character, you press the button twice, and so on.</p>
+
+<p>Given a string <code>s</code>, return <em>the <strong>minimum</strong> number of keypresses needed to type </em><code>s</code><em> using your keypad.</em></p>
+
+<p><strong>Note</strong> that the characters mapped to by each button, and the order they are mapped in cannot be changed.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2268.Minimum%20Number%20of%20Keypresses/images/image-20220505184346-1.png" style="width: 300px; height: 293px;" />
+<pre>
+<strong>Input:</strong> s = &quot;apple&quot;
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> One optimal way to setup your keypad is shown above.
+Type &#39;a&#39; by pressing button 1 once.
+Type &#39;p&#39; by pressing button 6 once.
+Type &#39;p&#39; by pressing button 6 once.
+Type &#39;l&#39; by pressing button 5 once.
+Type &#39;e&#39; by pressing button 3 once.
+A total of 5 button presses are needed, so return 5.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2268.Minimum%20Number%20of%20Keypresses/images/image-20220505203823-1.png" style="width: 300px; height: 288px;" />
+<pre>
+<strong>Input:</strong> s = &quot;abcdefghijkl&quot;
+<strong>Output:</strong> 15
+<strong>Explanation:</strong> One optimal way to setup your keypad is shown above.
+The letters &#39;a&#39; to &#39;i&#39; can each be typed by pressing a button once.
+Type &#39;j&#39; by pressing button 1 twice.
+Type &#39;k&#39; by pressing button 2 twice.
+Type &#39;l&#39; by pressing button 3 twice.
+A total of 15 button presses are needed, so return 15.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+</ul>
+
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2268.Minimum Number of Keypresses/README_EN.md

@@ -0,0 +1,86 @@
+# [2268. Minimum Number of Keypresses](https://leetcode.com/problems/minimum-number-of-keypresses)
+
+[中文文档](/solution/2200-2299/2268.Minimum%20Number%20of%20Keypresses/README.md)
+
+## Description
+
+<p>You have a keypad with <code>9</code> buttons, numbered from <code>1</code> to <code>9</code>, each mapped to lowercase English letters. You can choose which characters each button is matched to as long as:</p>
+
+<ul>
+	<li>All 26 lowercase English letters are mapped to.</li>
+	<li>Each character is mapped to by <strong>exactly</strong> <code>1</code> button.</li>
+	<li>Each button maps to <strong>at most</strong> <code>3</code> characters.</li>
+</ul>
+
+<p>To type the first character matched to a button, you press the button once. To type the second character, you press the button twice, and so on.</p>
+
+<p>Given a string <code>s</code>, return <em>the <strong>minimum</strong> number of keypresses needed to type </em><code>s</code><em> using your keypad.</em></p>
+
+<p><strong>Note</strong> that the characters mapped to by each button, and the order they are mapped in cannot be changed.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2268.Minimum%20Number%20of%20Keypresses/images/image-20220505184346-1.png" style="width: 300px; height: 293px;" />
+<pre>
+<strong>Input:</strong> s = &quot;apple&quot;
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> One optimal way to setup your keypad is shown above.
+Type &#39;a&#39; by pressing button 1 once.
+Type &#39;p&#39; by pressing button 6 once.
+Type &#39;p&#39; by pressing button 6 once.
+Type &#39;l&#39; by pressing button 5 once.
+Type &#39;e&#39; by pressing button 3 once.
+A total of 5 button presses are needed, so return 5.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2268.Minimum%20Number%20of%20Keypresses/images/image-20220505203823-1.png" style="width: 300px; height: 288px;" />
+<pre>
+<strong>Input:</strong> s = &quot;abcdefghijkl&quot;
+<strong>Output:</strong> 15
+<strong>Explanation:</strong> One optimal way to setup your keypad is shown above.
+The letters &#39;a&#39; to &#39;i&#39; can each be typed by pressing a button once.
+Type &#39;j&#39; by pressing button 1 twice.
+Type &#39;k&#39; by pressing button 2 twice.
+Type &#39;l&#39; by pressing button 3 twice.
+A total of 15 button presses are needed, so return 15.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2268.Minimum Number of Keypresses/images/image-20220505184346-1.png

@@ -0,0 +1,97 @@
+# [2269. 找到一个数字的 K 美丽值](https://leetcode.cn/problems/find-the-k-beauty-of-a-number)
+
+[English Version](/solution/2200-2299/2269.Find%20the%20K-Beauty%20of%20a%20Number/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>一个整数 <code>num</code>&nbsp;的&nbsp;<strong>k&nbsp;</strong>美丽值定义为&nbsp;<code>num</code>&nbsp;中符合以下条件的&nbsp;<strong>子字符串</strong>&nbsp;数目：</p>
+
+<ul>
+	<li>子字符串长度为&nbsp;<code>k</code>&nbsp;。</li>
+	<li>子字符串能整除 <code>num</code> 。</li>
+</ul>
+
+<p>给你整数&nbsp;<code>num</code> 和&nbsp;<code>k</code>&nbsp;，请你返回<em>&nbsp;</em><code>num</code>&nbsp;的 k 美丽值。</p>
+
+<p>注意：</p>
+
+<ul>
+	<li>允许有&nbsp;<strong>前缀</strong>&nbsp;<strong>0</strong>&nbsp;。</li>
+	<li><code>0</code>&nbsp;不能整除任何值。</li>
+</ul>
+
+<p>一个 <strong>子字符串</strong>&nbsp;是一个字符串里的连续一段字符序列。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>num = 240, k = 2
+<b>输出：</b>2
+<b>解释：</b>以下是 num 里长度为 k 的子字符串：
+- "<em><strong>24</strong></em>0" 中的 "24" ：24 能整除 240 。
+- "2<em><strong>40</strong></em>" 中的 "40" ：40 能整除 240 。
+所以，k 美丽值为 2 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>num = 430043, k = 2
+<b>输出：</b>2
+<b>解释：</b>以下是 num 里长度为 k 的子字符串：
+- "<em><strong>43</strong></em>0043" 中的 "43" ：43 能整除 430043 。
+- "4<em><strong>30</strong></em>043" 中的 "30" ：30 不能整除 430043 。
+- "43<em><strong>00</strong></em>43" 中的 "00" ：0 不能整除 430043 。
+- "430<em><strong>04</strong></em>3" 中的 "04" ：4 不能整除 430043 。
+- "4300<em><strong>43</strong></em>" 中的 "43" ：43 能整除 430043 。
+所以，k 美丽值为 2 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= num &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= k &lt;= num.length</code>&nbsp;（将&nbsp;<code>num</code>&nbsp;视为字符串）</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2268.Minimum Number of Keypresses/images/image-20220505203823-1.png

@@ -0,0 +1,87 @@
+# [2269. Find the K-Beauty of a Number](https://leetcode.com/problems/find-the-k-beauty-of-a-number)
+
+[中文文档](/solution/2200-2299/2269.Find%20the%20K-Beauty%20of%20a%20Number/README.md)
+
+## Description
+
+<p>The <strong>k-beauty</strong> of an integer <code>num</code> is defined as the number of <strong>substrings</strong> of <code>num</code> when it is read as a string that meet the following conditions:</p>
+
+<ul>
+	<li>It has a length of <code>k</code>.</li>
+	<li>It is a divisor of <code>num</code>.</li>
+</ul>
+
+<p>Given integers <code>num</code> and <code>k</code>, return <em>the k-beauty of </em><code>num</code>.</p>
+
+<p>Note:</p>
+
+<ul>
+	<li><strong>Leading zeros</strong> are allowed.</li>
+	<li><code>0</code> is not a divisor of any value.</li>
+</ul>
+
+<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = 240, k = 2
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The following are the substrings of num of length k:
+- &quot;24&quot; from &quot;<strong><u>24</u></strong>0&quot;: 24 is a divisor of 240.
+- &quot;40&quot; from &quot;2<u><strong>40</strong></u>&quot;: 40 is a divisor of 240.
+Therefore, the k-beauty is 2.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> num = 430043, k = 2
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The following are the substrings of num of length k:
+- &quot;43&quot; from &quot;<u><strong>43</strong></u>0043&quot;: 43 is a divisor of 430043.
+- &quot;30&quot; from &quot;4<u><strong>30</strong></u>043&quot;: 30 is not a divisor of 430043.
+- &quot;00&quot; from &quot;43<u><strong>00</strong></u>43&quot;: 0 is not a divisor of 430043.
+- &quot;04&quot; from &quot;430<u><strong>04</strong></u>3&quot;: 4 is not a divisor of 430043.
+- &quot;43&quot; from &quot;4300<u><strong>43</strong></u>&quot;: 43 is a divisor of 430043.
+Therefore, the k-beauty is 2.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= num &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= k &lt;= num.length</code> (taking <code>num</code> as a string)</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->