feat: update lc problems (#3125)

Author: yanglbme

solution/0100-0199/0130.Surrounded Regions/README.md

@@ -28,6 +28,8 @@ tags:
 	<li><strong>围绕：</strong>如果您可以用&nbsp;<code>'X'</code>&nbsp;单元格 <strong>连接这个区域</strong>，并且区域中没有任何单元格位于&nbsp;<code>board</code> 边缘，则该区域被 <code>'X'</code>&nbsp;单元格围绕。</li>
 </ul>
 
+<p>通过将输入矩阵&nbsp;<code>board</code> 中的所有 <code>'O'</code>&nbsp;替换为 <code>'X'</code> 来 <strong>捕获被围绕的区域</strong>。</p>
+
 <div class="original__bRMd">
 <div>
 <p>&nbsp;</p>

solution/0400-0499/0431.Encode N-ary Tree to Binary Tree/README.md

@@ -26,16 +26,47 @@ tags:
 
 <p>&nbsp;</p>
 
-<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0431.Encode%20N-ary%20Tree%20to%20Binary%20Tree/images/narytreebinarytreeexample.png" style="width: 500px;"></p>
+<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0431.Encode%20N-ary%20Tree%20to%20Binary%20Tree/images/narytreebinarytreeexample.png" style="width: 500px;" /></p>
 
 <p>&nbsp;</p>
 
+<pre>
+<strong>输入：</strong>root = [1,null,3,2,4,null,5,6]
+</pre>
+
 <p>注意，上面的方法仅仅是一个例子，可能可行也可能不可行。你没有必要遵循这种形式转化，你可以自己创造和实现不同的方法。</p>
 
-<p><strong>注意：</strong></p>
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>root = [1,null,3,2,4,null,5,6]
+<strong>输出：</strong>[1,null,3,2,4,null,5,6]
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+<strong>输出：</strong>[1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
+</pre>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>root = []
+<strong>输出：</strong>[]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
 
 <ol>
-	<li><code>N</code>&nbsp;的范围在 <code>[1, 1000]</code></li>
+	<li><code>N</code>&nbsp;的范围在 <code>[1, 10<sup>4</sup>]</code></li>
+	<li><code>0 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
+	<li>N 叉树的高度小于等于&nbsp;<code>1000</code>。</li>
 	<li>不要使用类成员 / 全局变量 / 静态变量来存储状态。你的编码和解码算法应是无状态的。</li>
 </ol>
 

solution/0400-0499/0469.Convex Polygon/README.md

@@ -4,6 +4,7 @@ difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0469.Convex%20Polygon/README.md
 tags:
     - 几何
+    - 数组
     - 数学
 ---
 

solution/0400-0499/0469.Convex Polygon/README_EN.md

@@ -4,6 +4,7 @@ difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0469.Convex%20Polygon/README_EN.md
 tags:
     - Geometry
+    - Array
     - Math
 ---
 

solution/0500-0599/0536.Construct Binary Tree from String/README.md

@@ -3,6 +3,7 @@ comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0536.Construct%20Binary%20Tree%20from%20String/README.md
 tags:
+    - 栈
     - 树
     - 深度优先搜索
     - 字符串

solution/0500-0599/0536.Construct Binary Tree from String/README_EN.md

@@ -3,6 +3,7 @@ comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0536.Construct%20Binary%20Tree%20from%20String/README_EN.md
 tags:
+    - Stack
     - Tree
     - Depth-First Search
     - String

solution/0500-0599/0558.Logical OR of Two Binary Grids Represented as Quad-Trees/README.md

@@ -35,11 +35,11 @@ tags:
 <pre>
 class Node {
     public boolean val;
-    public boolean isLeaf;
-    public Node topLeft;
-    public Node topRight;
-    public Node bottomLeft;
-    public Node bottomRight;
+&nbsp; &nbsp; public boolean isLeaf;
+&nbsp; &nbsp; public Node topLeft;
+&nbsp; &nbsp; public Node topRight;
+&nbsp; &nbsp; public Node bottomLeft;
+&nbsp; &nbsp; public Node bottomRight;
 }</pre>
 
 <p>我们可以按以下步骤为二维区域构建四叉树：</p>
@@ -60,9 +60,9 @@ class Node {
 
 <p>它与二叉树的序列化非常相似。唯一的区别是节点以列表形式表示 <code>[isLeaf, val]</code> 。</p>
 
-<p>如果 <code>isLeaf</code> 或者 <code>val</code> 的值为 True ，则表示它在列表 <code>[isLeaf, val]</code> 中的值为 <strong>1</strong> ；如果 <code>isLeaf</code> 或者 <code>val</code> 的值为 False ，则表示值为 <strong>0 </strong>。</p>
+<p>如果 <code>isLeaf</code> 或者 <code>val</code> 的值为 True ，则表示它在列表&nbsp;<code>[isLeaf, val]</code> 中的值为 <strong>1</strong> ；如果 <code>isLeaf</code> 或者 <code>val</code> 的值为 False ，则表示值为 <strong>0 </strong>。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
@@ -88,37 +88,13 @@ class Node {
 结果矩阵大小为 1*1，值全为 0 。
 </pre>
 
-<p><strong>示例 3：</strong></p>
-
-<pre>
-<strong>输入：</strong>quadTree1 = [[0,0],[1,0],[1,0],[1,1],[1,1]]
-, quadTree2 = [[0,0],[1,1],[1,1],[1,0],[1,1]]
-<strong>输出：</strong>[[1,1]]
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>quadTree1 = [[0,0],[1,1],[1,0],[1,1],[1,1]]
-, quadTree2 = [[0,0],[1,1],[0,1],[1,1],[1,1],null,null,null,null,[1,1],[1,0],[1,0],[1,1]]
-<strong>输出：</strong>[[0,0],[1,1],[0,1],[1,1],[1,1],null,null,null,null,[1,1],[1,0],[1,0],[1,1]]
-</pre>
-
-<p><strong>示例 5：</strong></p>
-
-<pre>
-<strong>输入：</strong>quadTree1 = [[0,1],[1,0],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
-, quadTree2 = [[0,1],[0,1],[1,0],[1,1],[1,0],[1,0],[1,0],[1,1],[1,1]]
-<strong>输出：</strong>[[0,0],[0,1],[0,1],[1,1],[1,0],[1,0],[1,0],[1,1],[1,1],[1,0],[1,0],[1,1],[1,1]]
-</pre>
-
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>quadTree1</code> 和 <code>quadTree2</code> 都是符合题目要求的四叉树，每个都代表一个 <code>n * n</code> 的矩阵。</li>
-	<li><code>n == 2^x</code> ，其中 <code>0 <= x <= 9</code>.</li>
+	<li><code>n == 2<sup>x</sup></code>&nbsp;，其中 <code>0 &lt;= x &lt;= 9</code>.</li>
 </ul>
 
 <!-- description:end -->

solution/1400-1499/1490.Clone N-ary Tree/README.md

@@ -58,7 +58,7 @@ class Node {
 
 <ul>
 	<li>给定的 N 叉树的深度小于或等于&nbsp;<code>1000</code>。</li>
-	<li>节点的总个数在&nbsp;<code>[0,&nbsp;10^4]</code>&nbsp;之间</li>
+	<li>节点的总个数在&nbsp;<code>[0,&nbsp;10<sup>4</sup>]</code>&nbsp;之间</li>
 </ul>
 
 <p>&nbsp;</p>

solution/1600-1699/1612.Check If Two Expression Trees are Equivalent/README.md

@@ -5,7 +5,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1612.Ch
 tags:
     - 树
     - 深度优先搜索
+    - 哈希表
     - 二叉树
+    - 计数
 ---
 
 <!-- problem:start -->

solution/1600-1699/1612.Check If Two Expression Trees are Equivalent/README_EN.md

@@ -5,7 +5,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1612.Ch
 tags:
     - Tree
     - Depth-First Search
+    - Hash Table
     - Binary Tree
+    - Counting
 ---
 
 <!-- problem:start -->

solution/1600-1699/1625.Lexicographically Smallest String After Applying Operations/README.md

@@ -5,8 +5,10 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1625.Le
 rating: 1992
 source: 第 211 场周赛 Q2
 tags:
+    - 深度优先搜索
     - 广度优先搜索
     - 字符串
+    - 枚举
 ---
 
 <!-- problem:start -->

solution/1600-1699/1625.Lexicographically Smallest String After Applying Operations/README_EN.md

@@ -5,8 +5,10 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1625.Le
 rating: 1992
 source: Weekly Contest 211 Q2
 tags:
+    - Depth-First Search
     - Breadth-First Search
     - String
+    - Enumeration
 ---
 
 <!-- problem:start -->

solution/1600-1699/1628.Design an Expression Tree With Evaluate Function/README.md

@@ -6,6 +6,7 @@ tags:
     - 栈
     - 树
     - 设计
+    - 数组
     - 数学
     - 二叉树
 ---

solution/1600-1699/1628.Design an Expression Tree With Evaluate Function/README_EN.md

@@ -6,6 +6,7 @@ tags:
     - Stack
     - Tree
     - Design
+    - Array
     - Math
     - Binary Tree
 ---

solution/1600-1699/1638.Count Substrings That Differ by One Character/README.md

@@ -8,6 +8,7 @@ tags:
     - 哈希表
     - 字符串
     - 动态规划
+    - 枚举
 ---
 
 <!-- problem:start -->

solution/1600-1699/1638.Count Substrings That Differ by One Character/README_EN.md

@@ -8,6 +8,7 @@ tags:
     - Hash Table
     - String
     - Dynamic Programming
+    - Enumeration
 ---
 
 <!-- problem:start -->

solution/2600-2699/2644.Find the Maximum Divisibility Score/README.md

@@ -37,13 +37,13 @@ tags:
 
 <p><code>divisors[0]</code>&nbsp;的可整除性分数为 2 因为&nbsp;<code>nums[2]</code> 和&nbsp;<code>nums[3]</code>&nbsp;能被 5 整除。</p>
 
-<p><code>divisors[1]</code> 的可整除性分数为 1 因为只有 <code>nums[1]</code>&nbsp;能被 3 整除。</p>
+<p><code>divisors[1]</code> 的可整除性分数为 2 因为&nbsp;<code>nums[1]</code>&nbsp;和&nbsp;<code>nums[2]</code>&nbsp;能被 3 整除。</p>
 
 <p><code>divisors[2]</code> 的可整除性分数为 0 因为&nbsp;<code>nums</code>&nbsp;中没有数字能被 7 整除。</p>
 
 <p><code>divisors[3]</code> 的可整除性分数为 2 因为 <code>nums[0]</code> 和&nbsp;<code>nums[3]</code>&nbsp;能够被 2 整除。</p>
 
-<p>因为&nbsp;<code>divisors[0]</code> 和&nbsp;<code>divisors[3]</code>&nbsp;有相同的可整除性分数，我们返回更小的那个&nbsp;<code>divisors[3]</code>。</p>
+<p>因为&nbsp;<code>divisors[0]</code>&nbsp;、<code>divisor[1]</code> 和&nbsp;<code>divisors[3]</code>&nbsp;有相同的可整除性分数，我们返回更小的那个&nbsp;<code>divisors[3]</code>。</p>
 </div>
 
 <p><strong class="example">示例 2：</strong></p>
@@ -71,13 +71,13 @@ tags:
 
 <p><strong>解释：</strong></p>
 
-<p><code>divisors[0]</code> 的可整除性分数为 0 因为&nbsp;<code>nums</code>&nbsp;中没有数字能被 10 整除。</p>
+<p><code>divisors[0]</code> 的可整除性分数为 2 因为&nbsp;<code>nums[0]</code>&nbsp;和&nbsp;<code>nums[3]</code> 能被 10 整除。</p>
 
 <p><code>divisors[1]</code> 的可整除性分数为 0 因为&nbsp;<code>nums</code>&nbsp;中没有数字能被 16&nbsp;整除。</p>
 
-<p><code>divisors[2]</code> 的可整除性分数为 0 因为&nbsp;<code>nums</code>&nbsp;中没有数字能被 20&nbsp;整除。</p>
+<p><code>divisors[2]</code> 的可整除性分数为 1 因为&nbsp;<code>nums[0]</code>&nbsp;能被 20&nbsp;整除。</p>
 
-<p>因为&nbsp;<code>divisors[0]</code>，<code>divisors[1]</code> 和&nbsp;<code>divisors[2]</code>&nbsp;都有相同的可整除性分数，我们返回最小的那个&nbsp;<code>divisors[0]</code>。</p>
+<p>因为&nbsp;<code>divisors[0]</code>&nbsp;的可整除性分数最大，我们返回&nbsp;<code>divisors[0]</code>。</p>
 </div>
 
 <p>&nbsp;</p>

solution/2600-2699/2644.Find the Maximum Divisibility Score/README_EN.md

@@ -36,13 +36,13 @@ tags:
 
 <p>The divisibility score of <code>divisors[0]</code> is 2 since <code>nums[2]</code> and <code>nums[3]</code> are divisible by 5.</p>
 
-<p>The divisibility score of <code>divisors[1]</code> is 1 since only <code>nums[1]</code> is divisible by 3.</p>
+<p>The divisibility score of <code>divisors[1]</code> is 2 since <code>nums[1]</code> and <code>nums[2]</code> are divisible by 3.</p>
 
 <p>The divisibility score of <code>divisors[2]</code> is 0 since none of the numbers in <code>nums</code> is divisible by 7.</p>
 
 <p>The divisibility score of <code>divisors[3]</code> is 2 since <code>nums[0]</code> and <code>nums[3]</code> are divisible by 2.</p>
 
-<p>As <code>divisors[0]</code> and <code>divisors[3]</code> have the same divisibility score, we return the smaller one which is <code>divisors[3]</code>.</p>
+<p>As <code>divisors[0]</code>,&nbsp;<code>divisors[1]</code>, and <code>divisors[3]</code> have the same divisibility score, we return the smaller one which is <code>divisors[3]</code>.</p>
 </div>
 
 <p><strong class="example">Example 2:</strong></p>
@@ -70,13 +70,11 @@ tags:
 
 <p><strong>Explanation:</strong></p>
 
-<p>The divisibility score of <code>divisors[0]</code> is 0 since none of the numbers in <code>nums</code> is divisible by 10.</p>
+<p>The divisibility score of <code>divisors[0]</code> is 2 since <code>nums[0]</code> and <code>nums[3]</code> are divisible by 10.</p>
 
 <p>The divisibility score of <code>divisors[1]</code> is 0 since none of the numbers in <code>nums</code> is divisible by 16.</p>
 
-<p>The divisibility score of <code>divisors[2]</code> is 0 since none of the numbers in <code>nums</code> is divisible by 20.</p>
-
-<p>As <code>divisors[0]</code>, <code>divisors[1]</code> and <code>divisors[2]</code> all have the same divisibility score, we return the smallest one which is <code>divisors[0]</code>.</p>
+<p>The divisibility score of <code>divisors[2]</code> is 1 since <code>nums[0]</code> is divisible by 20.</p>
 </div>
 
 <p>&nbsp;</p>

solution/2700-2799/2713.Maximum Strictly Increasing Cells in a Matrix/README.md

@@ -82,7 +82,17 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：排序 + 动态规划
+
+根据题目描述，我们顺序移动的单元格的值必须严格递增，因此，我们不妨用一个哈希表 $g$ 来记录每个值对应的所有单元格的位置，然后按照值的大小从小到大遍历。
+
+在这个过程中，我们可以维护两个数组 `rowMax` 和 `colMax`，分别记录每一行和每一列的最大递增长度。初始时，这两个数组的所有元素都为 $0$。
+
+对于每个值对应的所有单元格位置，我们按照位置顺序遍历，对于每个位置 $(i, j)$，我们可以计算出以该位置为终点的最大递增长度为 $1 + \max(\text{rowMax}[i], \text{colMax}[j])$，更新答案，然后更新 `rowMax[i]` 和 `colMax[j]`。
+
+最后返回答案即可。
+
+时间复杂度 $O(m \times n \times \log(m \times n))$，空间复杂度 $O(m \times n)$。
 
 <!-- tabs:start -->
 
@@ -212,6 +222,51 @@ func maxIncreasingCells(mat [][]int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxIncreasingCells(mat: number[][]): number {
+    const m = mat.length;
+    const n = mat[0].length;
+    const g: { [key: number]: [number, number][] } = {};
+
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            if (!g[mat[i][j]]) {
+                g[mat[i][j]] = [];
+            }
+            g[mat[i][j]].push([i, j]);
+        }
+    }
+
+    const rowMax = Array(m).fill(0);
+    const colMax = Array(n).fill(0);
+    let ans = 0;
+
+    const sortedKeys = Object.keys(g)
+        .map(Number)
+        .sort((a, b) => a - b);
+
+    for (const key of sortedKeys) {
+        const pos = g[key];
+        const mx: number[] = [];
+
+        for (const [i, j] of pos) {
+            mx.push(1 + Math.max(rowMax[i], colMax[j]));
+            ans = Math.max(ans, mx[mx.length - 1]);
+        }
+
+        for (let k = 0; k < pos.length; k++) {
+            const [i, j] = pos[k];
+            rowMax[i] = Math.max(rowMax[i], mx[k]);
+            colMax[j] = Math.max(colMax[j], mx[k]);
+        }
+    }
+
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->