feat: add new lc problems and solutions (doocs#1120)

Author: yanglbme

Diff for: .github/workflows/clang-format-lint.yml

@@ -19,9 +19,9 @@ jobs:
         inplace: True
 
     - uses: EndBug/add-and-commit@v9
-        with:
-          committer_name: yanglbme
-          committer_email: contact@yanglibin.info
-          message: 'chore: format code with clang-format'
-        env:
-          GITHUB_TOKEN: ${{ secrets.ACTION_TOKEN }}
+      with:
+        committer_name: yanglbme
+        committer_email: contact@yanglibin.info
+        message: 'chore: format code with clang-format'
+      env:
+        GITHUB_TOKEN: ${{ secrets.ACTION_TOKEN }}

Diff for: solution/0000-0099/0074.Search a 2D Matrix/README.md

@@ -6,14 +6,16 @@
 
 <!-- 这里写题目描述 -->
 
-<p>编写一个高效的算法来判断 <code>m x n</code> 矩阵中，是否存在一个目标值。该矩阵具有如下特性：</p>
+<p>给你一个满足下述两条属性的 <code>m x n</code> 整数矩阵：</p>
 
 <ul>
-	<li>每行中的整数从左到右按升序排列。</li>
+	<li>每行中的整数从左到右按非递减顺序排列。</li>
 	<li>每行的第一个整数大于前一行的最后一个整数。</li>
 </ul>
 
-<p> </p>
+<p>给你一个整数 <code>target</code> ，如果 <code>target</code> 在矩阵中，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>
+
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0074.Search%20a%202D%20Matrix/images/mat.jpg" style="width: 322px; height: 242px;" />
@@ -29,15 +31,15 @@
 <strong>输出：</strong>false
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>m == matrix.length</code></li>
 	<li><code>n == matrix[i].length</code></li>
-	<li><code>1 <= m, n <= 100</code></li>
-	<li><code>-10<sup>4</sup> <= matrix[i][j], target <= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= m, n &lt;= 100</code></li>
+	<li><code>-10<sup>4</sup> &lt;= matrix[i][j], target &lt;= 10<sup>4</sup></code></li>
 </ul>
 
 ## 解法

Diff for: solution/0300-0399/0373.Find K Pairs with Smallest Sums/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> sorted in <strong>ascending order</strong> and an integer <code>k</code>.</p>
+<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> sorted in <strong>non-decreasing&nbsp;order</strong> and an integer <code>k</code>.</p>
 
 <p>Define a pair <code>(u, v)</code> which consists of one element from the first array and one element from the second array.</p>
 
@@ -41,7 +41,7 @@
 <ul>
 	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>-10<sup>9</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
-	<li><code>nums1</code> and <code>nums2</code> both are sorted in <strong>ascending order</strong>.</li>
+	<li><code>nums1</code> and <code>nums2</code> both are sorted in <strong>non-decreasing order</strong>.</li>
 	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
 </ul>
 

Diff for: solution/1500-1599/1511.Customer Order Frequency/README.md

@@ -113,7 +113,6 @@ Winston 在2020年6月花费了300(300 * 1), 在7月花费了100(10 * 1 + 45 * 2
 Jonathan 在2020年6月花费了600(300 * 2), 在7月花费了20(2 * 10).
 Moustafa 在2020年6月花费了110 (10 * 2 + 45 * 2), 在7月花费了0.</pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

Diff for: solution/1500-1599/1511.Customer Order Frequency/README_EN.md

@@ -110,7 +110,6 @@ Jonathan spent 600 (300 * 2) in June and 20 ( 2 * 10) in July 2020.
 Moustafa spent 110 (10 * 2 + 45 * 2) in June and 0 in July 2020.
 </pre>
 
-
 ## Solutions
 
 <!-- tabs:start -->

Diff for: solution/2400-2499/2448.Minimum Cost to Make Array Equal/README_EN.md

@@ -45,6 +45,7 @@ It can be shown that we cannot make the array equal with a smaller cost.
 	<li><code>n == nums.length == cost.length</code></li>
 	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
 	<li><code>1 &lt;= nums[i], cost[i] &lt;= 10<sup>6</sup></code></li>
+	<li>Test cases are generated in a way that the output doesn&#39;t exceed&nbsp;2<sup>53</sup>-1</li>
 </ul>
 
 ## Solutions

Diff for: solution/2400-2499/2490.Circular Sentence/README.md

@@ -45,7 +45,7 @@
 <strong>输入：</strong>sentence = "eetcode"
 <strong>输出：</strong>true
 <strong>解释：</strong>句子中的单词是 ["eetcode"] 。
-- eetcod<em><strong>e</strong></em> 的最后一个字符和 eetcod<em><strong>e</strong></em> 的第一个字符相等。
+- eetcod<em><strong>e</strong></em> 的最后一个字符和 <em><strong>e</strong></em>etcod<em>e</em> 的第一个字符相等。
 这个句子是回环句。</pre>
 
 <p><strong>示例 3：</strong></p>
@@ -54,7 +54,7 @@
 <strong>输入：</strong>sentence = "Leetcode is cool"
 <strong>输出：</strong>false
 <strong>解释：</strong>句子中的单词是 ["Leetcode", "is", "cool"] 。
-- Leetcod<em><strong>e</strong></em>&nbsp;的最后一个字符和 i<strong><em>s</em></strong> 的第一个字符 <strong>不</strong> 相等。 
+- Leetcod<em><strong>e</strong></em>&nbsp;的最后一个字符和 <em><strong>i</strong></em>s 的第一个字符 <strong>不</strong> 相等。 
 这个句子 <strong>不</strong> 是回环句。</pre>
 
 <p>&nbsp;</p>

Diff for: solution/2600-2699/2627.Debounce/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>Given a function&nbsp;<code>fn</code> and a time in milliseconds&nbsp;<code>t</code>, return&nbsp;a&nbsp;<strong>debounced</strong>&nbsp;version of that function.</p>
 
-<p>A&nbsp;<strong>debounced</strong>&nbsp;function is a function whose execution is delayed by&nbsp;<code>t</code>&nbsp;milliseconds and whose&nbsp;execution is cancelled if it is called again within that window of time. The debounced function should also recieve the passed parameters.</p>
+<p>A&nbsp;<strong>debounced</strong>&nbsp;function is a function whose execution is delayed by&nbsp;<code>t</code>&nbsp;milliseconds and whose&nbsp;execution is cancelled if it is called again within that window of time. The debounced function should also receive the passed parameters.</p>
 
 <p>For example, let&#39;s say&nbsp;<code>t = 50ms</code>, and the function was called at&nbsp;<code>30ms</code>,&nbsp;<code>60ms</code>, and <code>100ms</code>. The first 2 function calls would be cancelled, and the 3rd function call would be executed at&nbsp;<code>150ms</code>. If instead&nbsp;<code>t = 35ms</code>, The 1st call would be cancelled, the 2nd would be executed at&nbsp;<code>95ms</code>, and the 3rd would be executed at&nbsp;<code>135ms</code>.</p>
 

Diff for: solution/2600-2699/2693.Call Function with Custom Context/README_EN.md

@@ -43,7 +43,7 @@ callPolyfill sets the "this" context to {"a": 5}. 7 is passe
 fn = function tax(price, taxRate) { 
  return `The cost of the ${this.item} is ${price * taxRate}`; 
 }
-args = [{"item": "burger"}, 10, 1,1]
+args = [{"item": "burger"}, 10, 1.1]
 <strong>Output:</strong> &quot;The cost of the burger is 11&quot;
 <strong>Explanation:</strong> callPolyfill sets the &quot;this&quot; context to {&quot;item&quot;: &quot;burger&quot;}. 10 and 1.1 are passed as additional arguments.
 </pre>

Diff for: solution/2600-2699/2695.Array Wrapper/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Create a class&nbsp;<code>ArrayWrapper</code> that accepts&nbsp;an array of integers in it&#39;s constructor. This class should have two features:</p>
+<p>Create a class&nbsp;<code>ArrayWrapper</code> that accepts&nbsp;an array of integers in its constructor. This class should have two features:</p>
 
 <ul>
 	<li>When two instances of this class are added together with the&nbsp;<code>+</code>&nbsp;operator, the resulting value is the sum of all the elements in&nbsp;both arrays.</li>

Diff for: solution/2700-2799/2753.Count Houses in a Circular Street II/README.md

@@ -0,0 +1,95 @@
+# [2753. 计算一个环形街道上的房屋数量 II](https://leetcode.cn/problems/count-houses-in-a-circular-street-ii)
+
+[English Version](/solution/2700-2799/2753.Count%20Houses%20in%20a%20Circular%20Street%20II/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给定一个代表 <strong>环形 </strong>街道的类 <code>Street</code> 和一个正整数 <code>k</code>，表示街道上房屋的最大数量（也就是说房屋数量不超过 <code>k</code>）。每个房屋的门初始时可以是开着的也可以是关着的（至少有一个房屋的门是开着的）。</p>
+
+<p>刚开始，你站在一座房子的门前。你的任务是计算街道上的房屋数量。</p>
+
+<p><code>Street</code> 类包含以下函数：</p>
+
+<ul>
+	<li><code>void closeDoor()</code>：关闭当前房屋的门。</li>
+	<li><code>boolean isDoorOpen()</code>：如果当前房屋的门是开着的返回 <code>true</code>，否则返回 <code>false</code>。</li>
+	<li><code>void moveRight()</code>：向右移动到下一座房屋。</li>
+</ul>
+
+<p><strong>注意：</strong>在<strong> 环形 </strong>街道内，如果将房屋从 <code>1</code> 到 <code>n</code> 编号，则当 <code>i &lt; n</code> 时&nbsp;<code>house<sub>i</sub></code> 右边的房子是&nbsp;<code>house<sub>i+1</sub></code>，<code>house<sub>n</sub></code> 右边的房子是&nbsp;<code>house<sub>1</sub></code>。</p>
+
+<p>返回 <code>ans</code>，它表示街道上的房屋数量。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>street = [1,1,1,1], k = 10
+<b>输出：</b>4
+<b>解释：</b>街道上有 4 座房屋，它们的门都是关着的。
+房屋数量小于 k，即 10。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>street = [1,0,1,1,0], k = 5
+<b>输出：</b>5
+<strong>解释：</strong>街道上有 5 座房屋，向右移动时第 1、3 和 4 座房屋的门是开着的，其余的门都是关着的。
+房屋数量等于 k，即 5。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n</code>&nbsp;是房屋数量</li>
+	<li><code>1 &lt;= n &lt;= k &lt;= 10<sup>5</sup></code></li>
+	<li><code>street</code>&nbsp;是环形的</li>
+	<li>输入数据中至少有一扇门是开着的</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
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

Diff for: solution/2700-2799/2753.Count Houses in a Circular Street II/README_EN.md

@@ -0,0 +1,85 @@
+# [2753. Count Houses in a Circular Street II](https://leetcode.com/problems/count-houses-in-a-circular-street-ii)
+
+[中文文档](/solution/2700-2799/2753.Count%20Houses%20in%20a%20Circular%20Street%20II/README.md)
+
+## Description
+
+<p>You are given an object <code>street</code> of class <code>Street</code> that represents a <strong>circular</strong> street and a positive integer <code>k</code> which represents a maximum bound for the number of houses in that street (in other words, the number of houses is less than or equal to <code>k</code>). Houses&#39; doors could be open or closed initially (at least one is open).</p>
+
+<p>Initially, you are standing in front of a door to a house on this street. Your task is to count the number of houses in the street.</p>
+
+<p>The class <code>Street</code> contains the following functions which may help you:</p>
+
+<ul>
+	<li><code>void closeDoor()</code>: Close the door of the house you are in front of.</li>
+	<li><code>boolean isDoorOpen()</code>: Returns <code>true</code> if the door of the current house is open and <code>false</code> otherwise.</li>
+	<li><code>void moveRight()</code>: Move to the right house.</li>
+</ul>
+
+<p><strong>Note</strong> that by <strong>circular</strong> street, we mean if you number the houses from <code>1</code> to <code>n</code>, then the right house of <code>house<sub>i</sub></code> is <code>house<sub>i+1</sub></code> for <code>i &lt; n</code>, and the right house of <code>house<sub>n</sub></code> is <code>house<sub>1</sub></code>.</p>
+
+<p>Return <code>ans</code> <em>which represents the number of houses on this street.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> street = [1,1,1,1], k = 10
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> There are 4 houses, and all their doors are open. 
+The number of houses is less than k, which is 10.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> street = [1,0,1,1,0], k = 5
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> There are 5 houses, and the doors of the 1st, 3rd, and 4th house (moving in the right direction) are open, and the rest are closed.
+The number of houses is equal to k, which is 5.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == number of houses</code></li>
+	<li><code>1 &lt;= n &lt;= k &lt;= 10<sup>5</sup></code></li>
+	<li><code>street</code> is circular by definition provided in the statement.</li>
+	<li>The input is generated such that at least one of the doors is open.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->