feat: add new lc problems

Author: yanglbme

solution/0000-0099/0005.Longest Palindromic Substring/README_EN.md

@@ -36,8 +36,8 @@
 
 Set $dp[i][j]$ to indicate whether the string $s[i..j]$ is a palindrome.
 
-- When $j - i \lt 2$, that is, the string length is `2`, as long as $s[i] == s[j]$, then $dp[i][j]$ is `true`.
-- When $j - i \ge 2$, there is $dp[i][j] = dp[i + 1][j - 1] \cap s[i] == s[j]$.
+-   When $j - i \lt 2$, that is, the string length is `2`, as long as $s[i] == s[j]$, then $dp[i][j]$ is `true`.
+-   When $j - i \ge 2$, there is $dp[i][j] = dp[i + 1][j - 1] \cap s[i] == s[j]$.
 
 The time complexity is $O(n^2)$ and the space complexity is $O(n^2)$. Where $n$ is the length of the string $s$.
 

solution/0000-0099/0026.Remove Duplicates from Sorted Array/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing order</strong>, remove the duplicates <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank"><strong>in-place</strong></a> such that each unique element appears only <strong>once</strong>. The <strong>relative order</strong> of the elements should be kept the <strong>same</strong>. Then return <em>the number of unique elements in </em><code>nums</code>.</p>
 
-<p>Consider the number of unique elements of <code>nums</code> be <code>k</code>, to get accepted, you need to do the following things:</p>
+<p>Consider the number of unique elements of <code>nums</code> to be <code>k</code>, to get accepted, you need to do the following things:</p>
 
 <ul>
 	<li>Change the array <code>nums</code> such that the first <code>k</code> elements of <code>nums</code> contain the unique elements in the order they were present in <code>nums</code> initially. The remaining elements of <code>nums</code> are not important as well as the size of <code>nums</code>.</li>

solution/0000-0099/0039.Combination Sum/README_EN.md

@@ -88,7 +88,7 @@ class Solution:
             t.append(candidates[i])
             dfs(i, s - candidates[i])
             t.pop()
-        
+
         candidates.sort()
         t = []
         ans = []

solution/0000-0099/0077.Combinations/README_EN.md

@@ -355,7 +355,7 @@ public class Solution {
     private List<int> t = new List<int>();
     private int n;
     private int k;
-    
+
     public IList<IList<int>> Combine(int n, int k) {
         this.n = n;
         this.k = k;
@@ -385,7 +385,7 @@ public class Solution {
     private List<int> t = new List<int>();
     private int n;
     private int k;
-    
+
     public IList<IList<int>> Combine(int n, int k) {
         this.n = n;
         this.k = k;

solution/0500-0599/0581.Shortest Unsorted Continuous Subarray/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Given an integer array <code>nums</code>, you need to find one <b>continuous subarray</b> that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order.</p>
+<p>Given an integer array <code>nums</code>, you need to find one <b>continuous subarray</b> such that if you only sort this subarray in non-decreasing order, then the whole array will be sorted in non-decreasing order.</p>
 
 <p>Return <em>the shortest such subarray and output its length</em>.</p>
 

solution/1000-1099/1023.Camelcase Matching/README.md

@@ -14,28 +14,31 @@
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入：</strong>queries = [&quot;FooBar&quot;,&quot;FooBarTest&quot;,&quot;FootBall&quot;,&quot;FrameBuffer&quot;,&quot;ForceFeedBack&quot;], pattern = &quot;FB&quot;
+<pre>
+<strong>输入：</strong>queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
 <strong>输出：</strong>[true,false,true,true,false]
 <strong>示例：</strong>
-&quot;FooBar&quot; 可以这样生成：&quot;F&quot; + &quot;oo&quot; + &quot;B&quot; + &quot;ar&quot;。
-&quot;FootBall&quot; 可以这样生成：&quot;F&quot; + &quot;oot&quot; + &quot;B&quot; + &quot;all&quot;.
-&quot;FrameBuffer&quot; 可以这样生成：&quot;F&quot; + &quot;rame&quot; + &quot;B&quot; + &quot;uffer&quot;.</pre>
+"FooBar" 可以这样生成："F" + "oo" + "B" + "ar"。
+"FootBall" 可以这样生成："F" + "oot" + "B" + "all".
+"FrameBuffer" 可以这样生成："F" + "rame" + "B" + "uffer".</pre>
 
 <p><strong>示例 2：</strong></p>
 
-<pre><strong>输入：</strong>queries = [&quot;FooBar&quot;,&quot;FooBarTest&quot;,&quot;FootBall&quot;,&quot;FrameBuffer&quot;,&quot;ForceFeedBack&quot;], pattern = &quot;FoBa&quot;
+<pre>
+<strong>输入：</strong>queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
 <strong>输出：</strong>[true,false,true,false,false]
 <strong>解释：</strong>
-&quot;FooBar&quot; 可以这样生成：&quot;Fo&quot; + &quot;o&quot; + &quot;Ba&quot; + &quot;r&quot;.
-&quot;FootBall&quot; 可以这样生成：&quot;Fo&quot; + &quot;ot&quot; + &quot;Ba&quot; + &quot;ll&quot;.
+"FooBar" 可以这样生成："Fo" + "o" + "Ba" + "r".
+"FootBall" 可以这样生成："Fo" + "ot" + "Ba" + "ll".
 </pre>
 
 <p><strong>示例 3：</strong></p>
 
-<pre><strong>输出：</strong>queries = [&quot;FooBar&quot;,&quot;FooBarTest&quot;,&quot;FootBall&quot;,&quot;FrameBuffer&quot;,&quot;ForceFeedBack&quot;], pattern = &quot;FoBaT&quot;
-<strong>输入：</strong>[false,true,false,false,false]
+<pre>
+<strong>输入：</strong>queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
+<strong>输出：</strong>[false,true,false,false,false]
 <strong>解释： </strong>
-&quot;FooBarTest&quot; 可以这样生成：&quot;Fo&quot; + &quot;o&quot; + &quot;Ba&quot; + &quot;r&quot; + &quot;T&quot; + &quot;est&quot;.
+"FooBarTest" 可以这样生成："Fo" + "o" + "Ba" + "r" + "T" + "est".
 </pre>
 
 <p>&nbsp;</p>

solution/1100-1199/1157.Online Majority Element In Subarray/README.md

@@ -54,7 +54,7 @@ majorityChecker.query(2,3,2); // 返回 2
 
 **方法一：线段树 + 摩尔投票 + 二分查找**
 
-我们注意到，题目需要我们找出特定区间内可能的众数，我们可以用线段树来维护每个区间内的候选众数以及其出现的次数。
+我们注意到，题目需要我们找出特定区间内可能的众数，考虑使用线段树来维护每个区间内的候选众数以及其出现的次数。
 
 我们定义线段树的每个节点为 `Node`，每个节点包含如下属性：
 

solution/1500-1599/1594.Maximum Non Negative Product in a Matrix/README.md

@@ -6,57 +6,47 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个大小为 <code>rows x cols</code> 的矩阵 <code>grid</code> 。最初，你位于左上角 <code>(0, 0)</code> ，每一步，你可以在矩阵中 <strong>向右</strong> 或 <strong>向下</strong> 移动。</p>
+<p>给你一个大小为 <code>m x n</code> 的矩阵 <code>grid</code> 。最初，你位于左上角 <code>(0, 0)</code> ，每一步，你可以在矩阵中 <strong>向右</strong> 或 <strong>向下</strong> 移动。</p>
 
-<p>在从左上角 <code>(0, 0)</code> 开始到右下角 <code>(rows - 1, cols - 1)</code> 结束的所有路径中，找出具有 <strong>最大非负积</strong> 的路径。路径的积是沿路径访问的单元格中所有整数的乘积。</p>
+<p>在从左上角 <code>(0, 0)</code> 开始到右下角 <code>(m - 1, n - 1)</code> 结束的所有路径中，找出具有 <strong>最大非负积</strong> 的路径。路径的积是沿路径访问的单元格中所有整数的乘积。</p>
 
-<p>返回 <strong>最大非负积 </strong>对<strong><em> </em><code>10<sup>9</sup>&nbsp;+ 7</code></strong> <strong>取余</strong> 的结果。如果最大积为负数，则返回<em> </em><code>-1</code> 。</p>
+<p>返回 <strong>最大非负积 </strong>对<strong><em> </em><code>10<sup>9</sup>&nbsp;+ 7</code></strong> <strong>取余</strong> 的结果。如果最大积为 <strong>负数</strong> ，则返回<em> </em><code>-1</code> 。</p>
 
 <p><strong>注意，</strong>取余是在得到最大积之后执行的。</p>
 
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[-1,-2,-3],
-&nbsp;            [-2,-3,-3],
-&nbsp;            [-3,-3,-2]]
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1594.Maximum%20Non%20Negative%20Product%20in%20a%20Matrix/images/product1.jpg" style="width: 244px; height: 245px;" />
+<pre>
+<strong>输入：</strong>grid = [[-1,-2,-3],[-2,-3,-3],[-3,-3,-2]]
 <strong>输出：</strong>-1
-<strong>解释：</strong>从 (0, 0) 到 (2, 2) 的路径中无法得到非负积，所以返回 -1
-</pre>
+<strong>解释：</strong>从 (0, 0) 到 (2, 2) 的路径中无法得到非负积，所以返回 -1 。</pre>
 
 <p><strong>示例 2：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[<strong>1</strong>,-2,1],
-&nbsp;            [<strong>1</strong>,<strong>-2</strong>,1],
-&nbsp;            [3,<strong>-4</strong>,<strong>1</strong>]]
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1594.Maximum%20Non%20Negative%20Product%20in%20a%20Matrix/images/product2.jpg" style="width: 244px; height: 245px;" />
+<pre>
+<strong>输入：</strong>grid = [[1,-2,1],[1,-2,1],[3,-4,1]]
 <strong>输出：</strong>8
-<strong>解释：</strong>最大非负积对应的路径已经用粗体标出 (1 * 1 * -2 * -4 * 1 = 8)
+<strong>解释：</strong>最大非负积对应的路径如图所示 (1 * 1 * -2 * -4 * 1 = 8)
 </pre>
 
 <p><strong>示例 3：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[<strong>1</strong>, 3],
-&nbsp;            [<strong>0</strong>,<strong>-4</strong>]]
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1594.Maximum%20Non%20Negative%20Product%20in%20a%20Matrix/images/product3.jpg" style="width: 164px; height: 165px;" />
+<pre>
+<strong>输入：</strong>grid = [[1,3],[0,-4]]
 <strong>输出：</strong>0
-<strong>解释：</strong>最大非负积对应的路径已经用粗体标出 (1 * 0 * -4 = 0)
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[ <strong>1</strong>, 4,4,0],
-&nbsp;            [<strong>-2</strong>, 0,0,1],
-&nbsp;            [ <strong>1</strong>,<strong>-1</strong>,<strong>1</strong>,<strong>1</strong>]]
-<strong>输出：</strong>2
-<strong>解释：</strong>最大非负积对应的路径已经用粗体标出 (1 * -2 * 1 * -1 * 1 * 1 = 2)
+<strong>解释：</strong>最大非负积对应的路径如图所示 (1 * 0 * -4 = 0)
 </pre>
 
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 &lt;= rows, cols &lt;= 15</code></li>
+	<li><code>m == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 15</code></li>
 	<li><code>-4 &lt;= grid[i][j] &lt;= 4</code></li>
 </ul>
 

solution/2500-2599/2581.Count Number of Possible Root Nodes/README_EN.md

@@ -72,7 +72,6 @@ Considering any node as root will give at least 1 correct guess.
 
 ## Solutions
 
-
 **Approach 1: Tree DP (change root)**
 
 First, we traverse the given edge set $edges$ and convert it to an adjacency list $g$ where $g[i]$ represents the adjacent nodes of node $i$. Use a hash map $gs$ to record the given guess set $guesses$.

solution/2600-2699/2605.Form Smallest Number From Two Digit Arrays/README_EN.md

@@ -34,7 +34,7 @@ Given two arrays of <strong>unique</strong> digits <code>nums1</code> and <code>
 
 ## Solutions
 
- **Approach 1: Enumeration**
+**Approach 1: Enumeration**
 
 We observe that if there are the same numbers in the arrays $nums1$ and $nums2$, then the minimum of the same numbers is the smallest number. Otherwise, we take the number $a$ in the array $nums1$ and the number $b$ in the array $nums2$, and concatenate the two numbers $a$ and $b$ into two numbers, and take the smaller number.
 
@@ -54,7 +54,7 @@ If the number $mask$ obtained by performing a bitwise AND operation on $mask1$ a
 
 Otherwise, we extract the position of the last $1$ in $mask1$ and $mask2$ respectively, and denote them as $a$ and $b$, respectively. Then the smallest number is $min(a \times 10 + b, b \times 10 + a)$.
 
-The time complexity is $O(m + n)$, and the space complexity is $O(1)$. Where $m$ and $n$ are the lengths of the arrays $nums1$ and $nums2$ respectively. 
+The time complexity is $O(m + n)$, and the space complexity is $O(1)$. Where $m$ and $n$ are the lengths of the arrays $nums1$ and $nums2$ respectively.
 
 <!-- tabs:start -->
 

solution/2600-2699/2612.Minimum Reverse Operations/README_EN.md

@@ -70,9 +70,9 @@ Therefore, for a specific index $i$, all its flipped indices form an arithmetic
 
 Next, we consider the range of values ​​of the index $i$ after flipping $j$.
 
-- If the boundary is not considered, the range of values ​​of $j$ is $[i - k + 1, i + k - 1]$.
-- If the subarray is on the left, then $[l, r] = [0, k - 1]$, so the flipped index $j$ of $i$ is $0 + k - 1 - i$, that is, $j = k - i - 1$, so the left boundary $mi = max(i - k + 1, k - i - 1)$.
-- If the subarray is on the right, then $[l, r] = [n - k, n - 1]$, so the flipped index $j= n - k + n - 1 - i$ is $j = n \times 2 - k - i - 1$, so the right boundary of $j$ is $mx = min(i + k - 1, n \times 2 - k - i - 1)$.
+-   If the boundary is not considered, the range of values ​​of $j$ is $[i - k + 1, i + k - 1]$.
+-   If the subarray is on the left, then $[l, r] = [0, k - 1]$, so the flipped index $j$ of $i$ is $0 + k - 1 - i$, that is, $j = k - i - 1$, so the left boundary $mi = max(i - k + 1, k - i - 1)$.
+-   If the subarray is on the right, then $[l, r] = [n - k, n - 1]$, so the flipped index $j= n - k + n - 1 - i$ is $j = n \times 2 - k - i - 1$, so the right boundary of $j$ is $mx = min(i + k - 1, n \times 2 - k - i - 1)$.
 
 We use two ordered sets to store all the odd indices and even indices to be searched, here we need to exclude the indices in the array $banned$ and the index $p$.
 

solution/2600-2699/2618.Check if Object Instance of Class/README.md

@@ -0,0 +1,70 @@
+# [2618. 检查是否是类的对象实例](https://leetcode.cn/problems/check-if-object-instance-of-class)
+
+[English Version](/solution/2600-2699/2618.Check%20if%20Object%20Instance%20of%20Class/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>请你编写一个函数，检查给定的对象是否是给定类或超类的实例。</p>
+
+<p>可以传递给函数的数据类型没有限制。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>func = () =&gt; checkIfInstance(new Date(), Date)
+<b>输出：</b>true
+<strong>解释：</strong>根据定义，Date 构造函数返回的对象是 Date 的一个实例。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>func = () =&gt; { class Animal {}; class Dog extends Animal {}; return checkIfInstance(new Dog(), Animal); }
+<b>输出：</b>true
+<strong>解释：</strong>
+class Animal {};
+class Dog extends Animal {};
+checkIfInstance(new Dog(), Animal); // true
+
+Dog 是 Animal 的子类。因此，Dog 对象同时是 Dog 和 Animal 的实例。</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<b>输入：</b>func = () =&gt; checkIfInstance(Date, Date)
+<b>输出：</b>false
+<strong>解释：</strong>日期的构造函数在逻辑上不能是其自身的实例。
+</pre>
+
+<p><strong>示例 4：</strong></p>
+
+<pre>
+<b>输入：</b>func = () =&gt; checkIfInstance(5, Number)
+<b>输出：</b>true
+<strong>解释：</strong>5 是一个 Number。注意，"instanceof" 关键字将返回 false。</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **TypeScript**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2600-2699/2618.Check if Object Instance of Class/README_EN.md

@@ -0,0 +1,64 @@
+# [2618. Check if Object Instance of Class](https://leetcode.com/problems/check-if-object-instance-of-class)
+
+[中文文档](/solution/2600-2699/2618.Check%20if%20Object%20Instance%20of%20Class/README.md)
+
+## Description
+
+<p>Write a function that checks if a given object is an instance of a given class or superclass.</p>
+
+<p>There are&nbsp;no constraints on the data types that can be passed to the function.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> func = () =&gt; checkIfInstance(new Date(), Date)
+<strong>Output:</strong> true
+<strong>Explanation: </strong>The object returned by the Date constructor is, by definition, an instance of Date.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> func = () =&gt; { class Animal {}; class Dog extends Animal {}; return checkIfInstance(new Dog(), Animal); }
+<strong>Output:</strong> true
+<strong>Explanation:</strong>
+class Animal {};
+class Dog extends Animal {};
+checkIfInstance(new Dog(), Animal); // true
+
+Dog is a subclass of Animal. Therefore, a Dog object is an instance of both Dog and Animal.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> func = () =&gt; checkIfInstance(Date, Date)
+<strong>Output:</strong> false
+<strong>Explanation: </strong>A date constructor cannot logically be an instance of itself.
+</pre>
+
+<p><strong class="example">Example 4:</strong></p>
+
+<pre>
+<strong>Input:</strong> func = () =&gt; checkIfInstance(5, Number)
+<strong>Output:</strong> true
+<strong>Explanation: </strong>5 is a Number. Note that the &quot;instanceof&quot; keyword would return false.
+</pre>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->