doocs · acbin · Aug 8, 2024 · Aug 8, 2024 · Aug 8, 2024
diff --git a/solution/0000-0099/0010.Regular Expression Matching/README.md b/solution/0000-0099/0010.Regular Expression Matching/README.md
@@ -25,7 +25,7 @@ tags:
 	<li><code>'*'</code> 匹配零个或多个前面的那一个元素</li>
 </ul>
 
-<p>所谓匹配，是要涵盖&nbsp;<strong>整个&nbsp;</strong>字符串&nbsp;<code>s</code>的，而不是部分字符串。</p>
+<p>所谓匹配，是要涵盖&nbsp;<strong>整个&nbsp;</strong>字符串&nbsp;<code>s</code> 的，而不是部分字符串。</p>
 &nbsp;
 
 <p><strong>示例 1：</strong></p>

diff --git a/solution/0000-0099/0075.Sort Colors/README.md b/solution/0000-0099/0075.Sort Colors/README.md
@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一个包含红色、白色和蓝色、共&nbsp;<code>n</code><em> </em>个元素的数组<meta charset="UTF-8" />&nbsp;<code>nums</code>&nbsp;，<strong><a href="https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank">原地</a></strong>对它们进行排序，使得相同颜色的元素相邻，并按照红色、白色、蓝色顺序排列。</p>
+<p>给定一个包含红色、白色和蓝色、共&nbsp;<code>n</code><em> </em>个元素的数组<meta charset="UTF-8" />&nbsp;<code>nums</code>&nbsp;，<strong><a href="https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank">原地</a>&nbsp;</strong>对它们进行排序，使得相同颜色的元素相邻，并按照红色、白色、蓝色顺序排列。</p>
 
 <p>我们使用整数 <code>0</code>、&nbsp;<code>1</code> 和 <code>2</code> 分别表示红色、白色和蓝色。</p>
 

diff --git a/solution/0000-0099/0085.Maximal Rectangle/README.md b/solution/0000-0099/0085.Maximal Rectangle/README.md
@@ -25,7 +25,7 @@ tags:
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0085.Maximal%20Rectangle/images/maximal.jpg" style="width: 402px; height: 322px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0085.Maximal%20Rectangle/images/1722912576-boIxpm-image.png" style="width: 402px; height: 322px;" />
 <pre>
 <strong>输入：</strong>matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
 <strong>输出：</strong>6

diff --git a/solution/0000-0099/0085.Maximal Rectangle/images/1722912576-boIxpm-image.png b/solution/0000-0099/0085.Maximal Rectangle/images/1722912576-boIxpm-image.png
diff --git a/solution/0200-0299/0277.Find the Celebrity/README.md b/solution/0200-0299/0277.Find the Celebrity/README.md
@@ -18,62 +18,56 @@ tags:
 
 <!-- description:start -->
 
-<p>假设你是一个专业的狗仔，参加了一个 <code>n</code> 人派对，其中每个人被从 <code>0</code> 到 <code>n - 1</code> 标号。在这个派对人群当中可能存在一位 “名人”。所谓 “名人” 的定义是：其他所有 <code>n - 1</code> 个人都认识他/她，而他/她并不认识其他任何人。</p>
+<p>假设你是一个专业的狗仔，参加了一个&nbsp;<code>n</code>&nbsp;人派对，其中每个人被从&nbsp;<code>0</code>&nbsp;到&nbsp;<code>n - 1</code>&nbsp;标号。在这个派对人群当中可能存在一位&nbsp;“名人”。所谓 “名人” 的定义是：其他所有&nbsp;<code>n - 1</code>&nbsp;个人都认识他/她，而他/她并不认识其他任何人。</p>
 
-<p>现在你想要确认这个 “名人” 是谁，或者确定这里没有 “名人”。而你唯一能做的就是问诸如 “A 你好呀，请问你认不认识 B呀？” 的问题，以确定 A 是否认识 B。你需要在（渐近意义上）尽可能少的问题内来确定这位 “名人” 是谁（或者确定这里没有 “名人”）。</p>
+<p>现在你想要确认这个 “名人” 是谁，或者确定这里没有&nbsp;“名人”。而你唯一能做的就是问诸如 “A&nbsp;你好呀，请问你认不认识&nbsp;B呀？”&nbsp;的问题，以确定 A 是否认识 B。你需要在（渐近意义上）尽可能少的问题内来确定这位 “名人” 是谁（或者确定这里没有 “名人”）。</p>
 
-<p>在本题中，你可以使用辅助函数 <code>bool knows(a, b)</code> 获取到 A 是否认识 B。请你来实现一个函数 <code>int findCelebrity(n)</code>。</p>
+<p>在本题中，你可以使用辅助函数&nbsp;<code>bool knows(a, b)</code>&nbsp;获取到 A&nbsp;是否认识 B。请你来实现一个函数&nbsp;<code>int findCelebrity(n)</code>。</p>
 
-<p>派对最多只会有一个 “名人” 参加。若 “名人” 存在，请返回他/她的编号；若 “名人” 不存在，请返回 <code>-1</code>。</p>
+<p>派对最多只会有一个 “名人” 参加。若&nbsp;“名人” 存在，请返回他/她的编号；若&nbsp;“名人”&nbsp;不存在，请返回&nbsp;<code>-1</code>。</p>
 
-<p> </p>
-
-<p><strong>示例 1:</strong></p>
-
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_1_bold.png" style="height: 181px; width: 186px;" /></p>
+<p>&nbsp;</p>
 
+<p><strong class="example">示例 1:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/g1.jpg" style="width: 224px; height: 145px;" />
 <pre>
-<strong>输入: </strong>graph = [
-  [1,1,0],
-  [0,1,0],
-  [1,1,1]
-]
+<strong>输入: </strong>graph = [[1,1,0],[0,1,0],[1,1,1]]
 <strong>输出: </strong>1
 <strong>解释: </strong>有编号分别为 0、1 和 2 的三个人。graph[i][j] = 1 代表编号为 i 的人认识编号为 j 的人，而 graph[i][j] = 0 则代表编号为 i 的人不认识编号为 j 的人。“名人” 是编号 1 的人，因为 0 和 2 均认识他/她，但 1 不认识任何人。
 </pre>
+<strong> </strong>
 
-<p><strong>示例 2:</strong></p>
-
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/277_example_2.png" style="height: 192px; width: 193px;" /></p>
-
+<p><strong><strong class="example">示例&nbsp;2:</strong></strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0277.Find%20the%20Celebrity/images/g2.jpg" style="width: 224px; height: 145px;" />
 <pre>
-<strong>输入: </strong>graph = [
-  [1,0,1],
-  [1,1,0],
-  [0,1,1]
-]
+<strong>输入: </strong>graph = [[1,0,1],[1,1,0],[0,1,1]]
 <strong>输出: </strong>-1
 <strong>解释: </strong>没有 “名人”
 </pre>
 
-<p> </p>
+<p><strong>&nbsp;</strong></p>
+<strong> </strong>
 
-<p><strong>提示：</strong></p>
+<p><strong><strong>提示：</strong></strong></p>
+<strong> </strong>
 
 <ul>
-	<li><code>n == graph.length</code></li>
-	<li><code>n == graph[i].length</code></li>
-	<li><code>2 <= n <= 100</code></li>
-	<li><code>graph[i][j]</code> 是 <code>0</code> 或 <code>1</code>.</li>
+	<li><code>n == graph.length == graph[i].length</code></li>
+	<li><code>2 &lt;= n &lt;= 100</code></li>
+	<li><code>graph[i][j]</code> 是 <code>0</code> 或 <code>1</code></li>
 	<li><code>graph[i][i] == 1</code></li>
 </ul>
+<strong> </strong>
 
-<p> </p>
+<p><strong>&nbsp;</strong></p>
+<strong> </strong>
 
-<p><strong>进阶：</strong>如果允许调用 API <code>knows</code> 的最大次数为 <code>3 * n</code> ，你可以设计一个不超过最大调用次数的解决方案吗？</p>
+<p><strong><strong>进阶：</strong></strong>如果允许调用 API <code>knows</code> 的最大次数为 <code>3 * n</code> ，你可以设计一个不超过最大调用次数的解决方案吗？</p>
+<strong> </strong>
 
 <ol>
 </ol>
+<strong> </strong>
 
 <!-- description:end -->
 

diff --git a/solution/0200-0299/0277.Find the Celebrity/images/277_example_1_bold.png b/solution/0200-0299/0277.Find the Celebrity/images/277_example_1_bold.png
diff --git a/solution/0200-0299/0277.Find the Celebrity/images/277_example_2.png b/solution/0200-0299/0277.Find the Celebrity/images/277_example_2.png
diff --git a/solution/0500-0599/0588.Design In-Memory File System/README.md b/solution/0500-0599/0588.Design In-Memory File System/README.md
@@ -7,6 +7,7 @@ tags:
     - 字典树
     - 哈希表
     - 字符串
+    - 排序
 ---
 
 <!-- problem:start -->

diff --git a/solution/0500-0599/0588.Design In-Memory File System/README_EN.md b/solution/0500-0599/0588.Design In-Memory File System/README_EN.md
@@ -7,6 +7,7 @@ tags:
     - Trie
     - Hash Table
     - String
+    - Sorting
 ---
 
 <!-- problem:start -->

diff --git a/solution/0600-0699/0642.Design Search Autocomplete System/README.md b/solution/0600-0699/0642.Design Search Autocomplete System/README.md
@@ -3,6 +3,7 @@ comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0642.Design%20Search%20Autocomplete%20System/README.md
 tags:
+    - 深度优先搜索
     - 设计
     - 字典树
     - 字符串

diff --git a/solution/0600-0699/0642.Design Search Autocomplete System/README_EN.md b/solution/0600-0699/0642.Design Search Autocomplete System/README_EN.md
@@ -3,6 +3,7 @@ comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0642.Design%20Search%20Autocomplete%20System/README_EN.md
 tags:
+    - Depth-First Search
     - Design
     - Trie
     - String

diff --git a/solution/0700-0799/0713.Subarray Product Less Than K/README.md b/solution/0700-0799/0713.Subarray Product Less Than K/README.md
@@ -26,7 +26,7 @@ tags:
 <pre>
 <strong>输入：</strong>nums = [10,5,2,6], k = 100
 <strong>输出：</strong>8
-<strong>解释：</strong>8 个乘积小于 100 的子数组分别为：[10]、[5]、[2],、[6]、[10,5]、[5,2]、[2,6]、[5,2,6]。
+<strong>解释：</strong>8 个乘积小于 100 的子数组分别为：[10]、[5]、[2]、[6]、[10,5]、[5,2]、[2,6]、[5,2,6]。
 需要注意的是 [10,5,2] 并不是乘积小于 100 的子数组。
 </pre>
 

diff --git a/solution/0800-0899/0840.Magic Squares In Grid/README.md b/solution/0800-0899/0840.Magic Squares In Grid/README.md
@@ -23,6 +23,8 @@ tags:
 
 <p>给定一个由整数组成的<code>row x col</code>&nbsp;的 <code>grid</code>，其中有多少个&nbsp;<code>3 × 3</code> 的 “幻方” 子矩阵？（每个子矩阵都是连续的）。</p>
 
+<p>注意：虽然幻方只能包含 1 到 9 的数字，但&nbsp;<code>grid</code> 可以包含最多15的数字。</p>
+
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>

diff --git a/solution/0800-0899/0840.Magic Squares In Grid/README_EN.md b/solution/0800-0899/0840.Magic Squares In Grid/README_EN.md
@@ -19,9 +19,11 @@ tags:
 
 <!-- description:start -->
 
-<p>A <code>3 x 3</code> magic square is a <code>3 x 3</code> grid filled with distinct numbers <strong>from </strong><code>1</code><strong> to </strong><code>9</code> such that each row, column, and both diagonals all have the same sum.</p>
+<p>A <code>3 x 3</code> <strong>magic square</strong> is a <code>3 x 3</code> grid filled with distinct numbers <strong>from </strong>1<strong> to </strong>9 such that each row, column, and both diagonals all have the same sum.</p>
 
-<p>Given a <code>row x col</code>&nbsp;<code>grid</code>&nbsp;of integers, how many <code>3 x 3</code> &quot;magic square&quot; subgrids are there?&nbsp; (Each subgrid is contiguous).</p>
+<p>Given a <code>row x col</code> <code>grid</code> of integers, how many <code>3 x 3</code> contiguous magic square subgrids are there?</p>
+
+<p>Note: while a magic square can only contain numbers from 1 to 9, <code>grid</code> may contain numbers up to 15.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

diff --git a/solution/0800-0899/0898.Bitwise ORs of Subarrays/README.md b/solution/0800-0899/0898.Bitwise ORs of Subarrays/README.md
@@ -18,11 +18,11 @@ tags:
 
 <!-- description:start -->
 
-<p>我们有一个非负整数数组<meta charset="UTF-8" />&nbsp;<code>arr</code>&nbsp;。</p>
+<p>给定一个整数数组<meta charset="UTF-8" />&nbsp;<code>arr</code>，返回所有&nbsp;<code>arr</code>&nbsp;的非空子数组的不同按位或的数量。</p>
 
-<p>对于每个（连续的）子数组<meta charset="UTF-8" />&nbsp;<code>sub = [arr[i], arr[i + 1], ..., arr[j]]</code>&nbsp;（&nbsp;<code>i &lt;= j</code>），我们对<meta charset="UTF-8" />&nbsp;<code>sub</code>&nbsp;中的每个元素进行按位或操作，获得结果<meta charset="UTF-8" />&nbsp;<code>arr[i] | arr[i + 1] | ... | arr[j]</code>&nbsp;。</p>
+<p>子数组的按位或是子数组中每个整数的按位或。含有一个整数的子数组的按位或就是该整数。</p>
 
-<p>返回可能结果的数量。 多次出现的结果在最终答案中仅计算一次。</p>
+<p><strong>子数组</strong> 是数组内连续的非空元素序列。</p>
 
 <p>&nbsp;</p>
 
@@ -61,7 +61,7 @@ tags:
 
 <ul>
 	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
-	<li><code>0 &lt;= nums[i]&nbsp;&lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= nums[i]&nbsp;&lt;= 10<sup>9</sup></code></li>
 </ul>
 
 <!-- description:end -->

diff --git a/solution/0900-0999/0930.Binary Subarrays With Sum/README_EN.md b/solution/0900-0999/0930.Binary Subarrays With Sum/README_EN.md
@@ -24,49 +24,32 @@ tags:
 <p>A <strong>subarray</strong> is a contiguous part of the array.</p>
 
 <p>&nbsp;</p>
-
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> nums = [1,0,1,0,1], goal = 2
-
 <strong>Output:</strong> 4
-
 <strong>Explanation:</strong> The 4 subarrays are bolded and underlined below:
-
 [<u><strong>1,0,1</strong></u>,0,1]
-
 [<u><strong>1,0,1,0</strong></u>,1]
-
 [1,<u><strong>0,1,0,1</strong></u>]
-
 [1,0,<u><strong>1,0,1</strong></u>]
-
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> nums = [0,0,0,0,0], goal = 0
-
 <strong>Output:</strong> 15
-
 </pre>
 
 <p>&nbsp;</p>
-
 <p><strong>Constraints:</strong></p>
 
 <ul>
-
-    <li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
-
-    <li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
-
-    <li><code>0 &lt;= goal &lt;= nums.length</code></li>
-
+	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
+	<li><code>0 &lt;= goal &lt;= nums.length</code></li>
 </ul>
 
 <!-- description:end -->

diff --git a/solution/1000-1099/1002.Find Common Characters/README.md b/solution/1000-1099/1002.Find Common Characters/README.md
@@ -20,7 +20,7 @@ tags:
 
 <!-- description:start -->
 
-给你一个字符串数组 <code>words</code> ，请你找出所有在 <code>words</code> 的每个字符串中都出现的共用字符（ <strong>包括重复字符</strong>），并以数组形式返回。你可以按 <strong>任意顺序</strong> 返回答案。
+给你一个字符串数组 <code>words</code> ，请你找出所有在 <code>words</code> 的每个字符串中都出现的共用字符（<strong>包括重复字符</strong>），并以数组形式返回。你可以按 <strong>任意顺序</strong> 返回答案。
 
 <p>&nbsp;</p>
 

diff --git a/solution/1300-1399/1395.Count Number of Teams/README.md b/solution/1300-1399/1395.Count Number of Teams/README.md
@@ -6,6 +6,7 @@ rating: 1343
 source: 第 182 场周赛 Q2
 tags:
     - 树状数组
+    - 线段树
     - 数组
     - 动态规划
 ---

diff --git a/solution/1300-1399/1395.Count Number of Teams/README_EN.md b/solution/1300-1399/1395.Count Number of Teams/README_EN.md
@@ -6,6 +6,7 @@ rating: 1343
 source: Weekly Contest 182 Q2
 tags:
     - Binary Indexed Tree
+    - Segment Tree
     - Array
     - Dynamic Programming
 ---

diff --git a/solution/1400-1499/1444.Number of Ways of Cutting a Pizza/README.md b/solution/1400-1499/1444.Number of Ways of Cutting a Pizza/README.md
@@ -73,7 +73,7 @@ tags:
 我们可以使用二维前缀和来快速计算出每个子矩形中苹果的数量，定义 $s[i][j]$ 表示矩形前 $i$ 行，前 $j$ 列的子矩形中苹果的数量，那么 $s[i][j]$ 可以由 $s[i-1][j]$, $s[i][j-1]$, $s[i-1][j-1]$ 三个子矩形的苹果数量求得，具体的计算方法如下：
 
 $$
-s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + (pizza[i-1][j-1] == 'A')
+s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + \textit{int}(pizza[i-1][j-1] == 'A')
 $$
 
 其中 $pizza[i-1][j-1]$ 表示矩形中第 $i$ 行，第 $j$ 列的字符，如果是苹果，则为 $1$，否则为 $0$。

diff --git a/solution/1400-1499/1444.Number of Ways of Cutting a Pizza/README_EN.md b/solution/1400-1499/1444.Number of Ways of Cutting a Pizza/README_EN.md
@@ -69,7 +69,30 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: 2D Prefix Sum + Memoized Search
+
+We can use a 2D prefix sum to quickly calculate the number of apples in each sub-rectangle. Define $s[i][j]$ to represent the number of apples in the sub-rectangle that includes the first $i$ rows and the first $j$ columns. Then $s[i][j]$ can be derived from the number of apples in the three sub-rectangles $s[i-1][j]$, $s[i][j-1]$, and $s[i-1][j-1]$. The specific calculation method is as follows:
+
+$$
+s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + (pizza[i-1][j-1] == 'A')
+$$
+
+Here, $pizza[i-1][j-1]$ represents the character at the $i$-th row and $j$-th column in the rectangle. If it is an apple, it is $1$; otherwise, it is $0$.
+
+Next, we design a function $dfs(i, j, k)$, which represents the number of ways to cut the rectangle $(i, j, m-1, n-1)$ with $k$ cuts to get $k+1$ pieces of pizza. Here, $(i, j)$ and $(m-1, n-1)$ are the coordinates of the top-left and bottom-right corners of the rectangle, respectively. The calculation method of the function $dfs(i, j, k)$ is as follows:
+
+-   If $k = 0$, it means no more cuts can be made. We need to check if there are any apples in the rectangle. If there are apples, return $1$; otherwise, return $0$.
+-   If $k \gt 0$, we need to enumerate the position of the last cut. If the last cut is horizontal, we need to enumerate the cutting position $x$, where $i \lt x \lt m$. If $s[x][n] - s[i][n] - s[x][j] + s[i][j] \gt 0$, it means there are apples in the upper piece of pizza, and we add the value of $dfs(x, j, k-1)$ to the answer. If the last cut is vertical, we need to enumerate the cutting position $y$, where $j \lt y \lt n$. If $s[m][y] - s[i][y] - s[m][j] + s[i][j] \gt 0$, it means there are apples in the left piece of pizza, and we add the value of $dfs(i, y, k-1)$ to the answer.
+
+The final answer is the value of $dfs(0, 0, k-1)$.
+
+To avoid repeated calculations, we can use memoized search. We use a 3D array $f$ to record the value of $dfs(i, j, k)$. When we need to calculate the value of $dfs(i, j, k)$, if $f[i][j][k]$ is not $-1$, it means we have already calculated it before, and we can directly return $f[i][j][k]$. Otherwise, we calculate the value of $dfs(i, j, k)$ according to the above method and save the result in $f[i][j][k]$.
+
+The time complexity is $O(m \times n \times k \times (m + n))$, and the space complexity is $O(m \times n \times k)$. Here, $m$ and $n$ are the number of rows and columns of the rectangle, respectively.
+
+Similar problems:
+
+-   [2312. Selling Pieces of Wood](https://github.com/doocs/leetcode/blob/main/solution/2300-2399/2312.Selling%20Pieces%20of%20Wood/README_EN.md)
 
 <!-- tabs:start -->
 

diff --git a/solution/1400-1499/1446.Consecutive Characters/README.md b/solution/1400-1499/1446.Consecutive Characters/README.md
@@ -57,11 +57,11 @@ tags:
 
 ### 方法一：遍历计数
 
-我们定义一个变量 $t$，表示当前连续字符的长度，初始时 $t=1$。
+我们定义一个变量 $\textit{t}$，表示当前连续字符的长度，初始时 $\textit{t}=1$。
 
-接下来，我们从字符串 $s$ 的第二个字符开始遍历，如果当前字符与上一个字符相同，那么 $t = t + 1$，然后更新答案 $ans = \max(ans, t)$；否则，$t = 1$。
+接下来，我们从字符串 $s$ 的第二个字符开始遍历，如果当前字符与上一个字符相同，那么 $\textit{t} = \textit{t} + 1$，然后更新答案 $\textit{ans} = \max(\textit{ans}, \textit{t})$；否则 $\textit{t} = 1$。
 
-最后返回答案 $ans$ 即可。
+最后返回答案 $\textit{ans}$ 即可。
 
 时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$。
 

diff --git a/solution/1400-1499/1446.Consecutive Characters/README_EN.md b/solution/1400-1499/1446.Consecutive Characters/README_EN.md
@@ -53,7 +53,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Traversal and Counting
+
+We define a variable $\textit{t}$ to represent the length of the current consecutive characters, initially $\textit{t}=1$.
+
+Next, we traverse the string $s$ starting from the second character. If the current character is the same as the previous character, then $\textit{t} = \textit{t} + 1$, and update the answer $\textit{ans} = \max(\textit{ans}, \textit{t})$; otherwise, set $\textit{t} = 1$.
+
+Finally, return the answer $\textit{ans}$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

diff --git a/solution/1800-1899/1820.Maximum Number of Accepted Invitations/README.md b/solution/1800-1899/1820.Maximum Number of Accepted Invitations/README.md
@@ -3,6 +3,7 @@ comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/1800-1899/1820.Maximum%20Number%20of%20Accepted%20Invitations/README.md
 tags:
+    - 图
     - 数组
     - 回溯
     - 矩阵

diff --git a/solution/1800-1899/1820.Maximum Number of Accepted Invitations/README_EN.md b/solution/1800-1899/1820.Maximum Number of Accepted Invitations/README_EN.md
@@ -3,6 +3,7 @@ comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/1800-1899/1820.Maximum%20Number%20of%20Accepted%20Invitations/README_EN.md
 tags:
+    - Graph
     - Array
     - Backtracking
     - Matrix
-Original file line number
+Diff line change
@@ Expand Up / @@ -7,6 +7,7 @@ tags: @@
         - Trie
         - Hash Table
         - String
+        - Sorting
     ---
     <!-- problem:start -->
@@ Expand Down @@