doocs · acbin · Jan 29, 2024 · Jan 29, 2024
diff --git a/solution/0100-0199/0197.Rising Temperature/README.md b/solution/0100-0199/0197.Rising Temperature/README.md
@@ -19,6 +19,7 @@
 | temperature   | int     |
 +---------------+---------+
 id 是该表具有唯一值的列。
+没有具有相同 recordDate 的不同行。
 该表包含特定日期的温度信息</pre>
 
 <p>&nbsp;</p>

diff --git a/solution/0100-0199/0197.Rising Temperature/README_EN.md b/solution/0100-0199/0197.Rising Temperature/README_EN.md
@@ -15,6 +15,7 @@
 | temperature   | int     |
 +---------------+---------+
 id is the column with unique values for this table.
+There are no different rows with the same recordDate.
 This table contains information about the temperature on a certain day.
 </pre>
 

diff --git a/solution/0500-0599/0521.Longest Uncommon Subsequence I/README_EN.md b/solution/0500-0599/0521.Longest Uncommon Subsequence I/README_EN.md
@@ -4,15 +4,9 @@
 
 ## Description
 
-<p>Given two strings <code>a</code> and <code>b</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between </em><code>a</code> <em>and</em> <code>b</code>. If the longest uncommon subsequence does not exist, return <code>-1</code>.</p>
+<p>Given two strings <code>a</code> and <code>b</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between </em><code>a</code> <em>and</em> <code>b</code>. <em>If no such uncommon subsequence exists, return</em> <code>-1</code><em>.</em></p>
 
-<p>An <strong>uncommon subsequence</strong> between two strings is a string that is a <strong>subsequence of one but not the other</strong>.</p>
-
-<p>A <strong>subsequence</strong> of a string <code>s</code> is a string that can be obtained after deleting any number of characters from <code>s</code>.</p>
-
-<ul>
-	<li>For example, <code>&quot;abc&quot;</code> is a subsequence of <code>&quot;aebdc&quot;</code> because you can delete the underlined characters in <code>&quot;a<u>e</u>b<u>d</u>c&quot;</code> to get <code>&quot;abc&quot;</code>. Other subsequences of <code>&quot;aebdc&quot;</code> include <code>&quot;aebdc&quot;</code>, <code>&quot;aeb&quot;</code>, and <code>&quot;&quot;</code> (empty string).</li>
-</ul>
+<p>An <strong>uncommon subsequence</strong> between two strings is a string that is a <strong><span data-keyword="subsequence-string">subsequence</span> of exactly one of them</strong>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
@@ -37,7 +31,7 @@ Note that &quot;cdc&quot; is also a longest uncommon subsequence.
 <pre>
 <strong>Input:</strong> a = &quot;aaa&quot;, b = &quot;aaa&quot;
 <strong>Output:</strong> -1
-<strong>Explanation:</strong>&nbsp;Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a.
+<strong>Explanation:</strong>&nbsp;Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a. So the answer would be <code>-1</code>.
 </pre>
 
 <p>&nbsp;</p>

diff --git a/solution/0600-0699/0606.Construct String from Binary Tree/README_EN.md b/solution/0600-0699/0606.Construct String from Binary Tree/README_EN.md
@@ -4,25 +4,43 @@
 
 ## Description
 
-<p>Given the <code>root</code> of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.</p>
+<p>Given the <code>root</code> node of a binary tree, your task is to create a string representation of the tree following a specific set of formatting rules. The representation should be based on a preorder traversal of the binary tree and must adhere to the following guidelines:</p>
 
-<p>Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.</p>
+<ul>
+	<li>
+	<p><strong>Node Representation</strong>: Each node in the tree should be represented by its integer value.</p>
+	</li>
+	<li>
+	<p><strong>Parentheses for Children</strong>: If a node has at least one child (either left or right), its children should be represented inside parentheses. Specifically:</p>
+
+    <ul>
+    	<li>If a node has a left child, the value of the left child should be enclosed in parentheses immediately following the node&#39;s value.</li>
+    	<li>If a node has a right child, the value of the right child should also be enclosed in parentheses. The parentheses for the right child should follow those of the left child.</li>
+    </ul>
+    </li>
+    <li>
+    <p><strong>Omitting Empty Parentheses</strong>: Any empty parentheses pairs (i.e., <code>()</code>) should be omitted from the final string representation of the tree, with one specific exception: when a node has a right child but no left child. In such cases, you must include an empty pair of parentheses to indicate the absence of the left child. This ensures that the one-to-one mapping between the string representation and the original binary tree structure is maintained.</p>
+
+    <p>In summary, empty parentheses pairs should be omitted when a node has only a left child or no children. However, when a node has a right child but no left child, an empty pair of parentheses must precede the representation of the right child to reflect the tree&#39;s structure accurately.</p>
+    </li>
+
+</ul>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons1-tree.jpg" style="width: 292px; height: 301px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons1-tree.jpg" style="padding: 10px; background: #fff; border-radius: .5rem;" />
 <pre>
 <strong>Input:</strong> root = [1,2,3,4]
 <strong>Output:</strong> &quot;1(2(4))(3)&quot;
-<strong>Explanation:</strong> Originally, it needs to be &quot;1(2(4)())(3()())&quot;, but you need to omit all the unnecessary empty parenthesis pairs. And it will be &quot;1(2(4))(3)&quot;
+<strong>Explanation:</strong> Originally, it needs to be &quot;1(2(4)())(3()())&quot;, but you need to omit all the empty parenthesis pairs. And it will be &quot;1(2(4))(3)&quot;.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons2-tree.jpg" style="width: 207px; height: 293px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0606.Construct%20String%20from%20Binary%20Tree/images/cons2-tree.jpg" style="padding: 10px; background: #fff; border-radius: .5rem;" />
 <pre>
 <strong>Input:</strong> root = [1,2,3,null,4]
 <strong>Output:</strong> &quot;1(2()(4))(3)&quot;
-<strong>Explanation:</strong> Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
+<strong>Explanation:</strong> Almost the same as the first example, except the <code>()</code> after <code>2</code> is necessary to indicate the absence of a left child for <code>2</code> and the presence of a right child.
 </pre>
 
 <p>&nbsp;</p>

diff --git a/solution/0800-0899/0827.Making A Large Island/README.md b/solution/0800-0899/0827.Making A Large Island/README.md
@@ -6,44 +6,44 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个大小为 <code>n x n</code> 二进制矩阵 <code>grid</code> 。<strong>最多</strong> 只能将一格 <code>0</code> 变成 <code>1</code> 。</p>
+<p>给你一个大小为 <code>n x n</code> 二进制矩阵 <code>grid</code> 。<strong>最多</strong> 只能将一格&nbsp;<code>0</code> 变成&nbsp;<code>1</code> 。</p>
 
 <p>返回执行此操作后，<code>grid</code> 中最大的岛屿面积是多少？</p>
 
-<p><strong>岛屿</strong> 由一组上、下、左、右四个方向相连的 <code>1</code> 形成。</p>
+<p><strong>岛屿</strong> 由一组上、下、左、右四个方向相连的&nbsp;<code>1</code> 形成。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
-<p><strong>示例 1:</strong></p>
+<p><strong class="example">示例 1:</strong></p>
 
 <pre>
 <strong>输入: </strong>grid = [[1, 0], [0, 1]]
 <strong>输出:</strong> 3
 <strong>解释:</strong> 将一格0变成1，最终连通两个小岛得到面积为 3 的岛屿。
 </pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong class="example">示例 2:</strong></p>
 
 <pre>
 <strong>输入: </strong>grid =<strong> </strong>[[1, 1], [1, 0]]
 <strong>输出:</strong> 4
 <strong>解释:</strong> 将一格0变成1，岛屿的面积扩大为 4。</pre>
 
-<p><strong>示例 3:</strong></p>
+<p><strong class="example">示例 3:</strong></p>
 
 <pre>
 <strong>输入: </strong>grid = [[1, 1], [1, 1]]
 <strong>输出:</strong> 4
 <strong>解释:</strong> 没有0可以让我们变成1，面积依然为 4。</pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>n == grid.length</code></li>
 	<li><code>n == grid[i].length</code></li>
-	<li><code>1 <= n <= 500</code></li>
+	<li><code>1 &lt;= n &lt;= 500</code></li>
 	<li><code>grid[i][j]</code> 为 <code>0</code> 或 <code>1</code></li>
 </ul>
 

diff --git a/solution/1600-1699/1657.Determine if Two Strings Are Close/README.md b/solution/1600-1699/1657.Determine if Two Strings Are Close/README.md
@@ -12,12 +12,12 @@
 	<li>操作 1：交换任意两个 <strong>现有</strong> 字符。
 
     <ul>
-    	<li>例如，<code>a<strong>b</strong>cd<strong>e</strong> -> a<strong>e</strong>cd<strong>b</strong></code></li>
+    	<li>例如，<code>a<u>b</u>cd<u>e</u> -&gt; a<u>e</u>cd<u>b</u></code></li>
     </ul>
     </li>
     <li>操作 2：将一个 <strong>现有</strong> 字符的每次出现转换为另一个 <strong>现有</strong> 字符，并对另一个字符执行相同的操作。
     <ul>
-    	<li>例如，<code><strong>aa</strong>c<strong>abb</strong> -> <strong>bb</strong>c<strong>baa</strong></code>（所有 <code>a</code> 转化为 <code>b</code> ，而所有的 <code>b</code> 转换为 <code>a</code> ）</li>
+    	<li>例如，<code><u>aa</u>c<u>abb</u> -&gt; <u>bb</u>c<u>baa</u></code>（所有 <code>a</code> 转化为 <code>b</code> ，而所有的 <code>b</code> 转换为 <code>a</code> ）</li>
     </ul>
     </li>
 
@@ -27,16 +27,16 @@
 
 <p>给你两个字符串，<code>word1</code> 和 <code>word2</code> 。如果<em> </em><code>word1</code><em> </em>和<em> </em><code>word2</code><em> </em><strong>接近 </strong>，就返回 <code>true</code> ；否则，返回<em> </em><code>false</code><em> </em>。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
 <pre>
 <strong>输入：</strong>word1 = "abc", word2 = "bca"
 <strong>输出：</strong>true
 <strong>解释：</strong>2 次操作从 word1 获得 word2 。
-执行操作 1："a<strong>bc</strong>" -> "a<strong>cb</strong>"
-执行操作 1："<strong>a</strong>c<strong>b</strong>" -> "<strong>b</strong>c<strong>a</strong>"
+执行操作 1："a<u>bc</u>" -&gt; "a<u>cb</u>"
+执行操作 1："<u>a</u>c<u>b</u>" -&gt; "<u>b</u>c<u>a</u>"
 </pre>
 
 <p><strong>示例 2：</strong></p>
@@ -52,24 +52,15 @@
 <strong>输入：</strong>word1 = "cabbba", word2 = "abbccc"
 <strong>输出：</strong>true
 <strong>解释：</strong>3 次操作从 word1 获得 word2 。
-执行操作 1："ca<strong>b</strong>bb<strong>a</strong>" -> "ca<strong>a</strong>bb<strong>b</strong>"
-执行操作 2：<code>"</code><strong>c</strong>aa<strong>bbb</strong>" -> "<strong>b</strong>aa<strong>ccc</strong>"
-执行操作 2："<strong>baa</strong>ccc" -> "<strong>abb</strong>ccc"
+执行操作 1："ca<u>b</u>bb<u>a</u>" -&gt; "ca<u>a</u>bb<u>b</u>"
+执行操作 2：<code>"</code><u>c</u>aa<u>bbb</u>" -&gt; "<u>b</u>aa<u>ccc</u>"
+执行操作 2："<u>baa</u>ccc" -&gt; "<u>abb</u>ccc"
 </pre>
 
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>word1 = "cabbba", word2 = "aabbss"
-<strong>输出：</strong>false
-<strong>解释：</strong>不管执行多少次操作，都无法从 word1 得到 word2 ，反之亦然。</pre>
-
-<p> </p>
-
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= word1.length, word2.length <= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= word1.length, word2.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>word1</code> 和 <code>word2</code> 仅包含小写英文字母</li>
 </ul>
 

diff --git a/solution/1600-1699/1657.Determine if Two Strings Are Close/README_EN.md b/solution/1600-1699/1657.Determine if Two Strings Are Close/README_EN.md
@@ -51,7 +51,7 @@ Apply Operation 1: &quot;<u>a</u>c<u>b</u>&quot; -&gt; &quot;<u>b</u>c<u>a</u>&q
 <strong>Output:</strong> true
 <strong>Explanation:</strong> You can attain word2 from word1 in 3 operations.
 Apply Operation 1: &quot;ca<u>b</u>bb<u>a</u>&quot; -&gt; &quot;ca<u>a</u>bb<u>b</u>&quot;
-<code>Apply Operation 2: &quot;</code><u>c</u>aa<u>bbb</u>&quot; -&gt; &quot;<u>b</u>aa<u>ccc</u>&quot;
+Apply Operation 2: &quot;<u>c</u>aa<u>bbb</u>&quot; -&gt; &quot;<u>b</u>aa<u>ccc</u>&quot;
 Apply Operation 2: &quot;<u>baa</u>ccc&quot; -&gt; &quot;<u>abb</u>ccc&quot;
 </pre>
 

diff --git a/...2100-2199/2186.Minimum Number of Steps to Make Two Strings Anagram II/README.md b/...2100-2199/2186.Minimum Number of Steps to Make Two Strings Anagram II/README.md
@@ -1,4 +1,4 @@
-# [2186. 使两字符串互为字母异位词的最少步骤数](https://leetcode.cn/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii)
+# [2186. 制造字母异位词的最小步骤数 II](https://leetcode.cn/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii)
 
 [English Version](/solution/2100-2199/2186.Minimum%20Number%20of%20Steps%20to%20Make%20Two%20Strings%20Anagram%20II/README_EN.md)
 

diff --git a/solution/2200-2299/2254.Design Video Sharing Platform/README_EN.md b/solution/2200-2299/2254.Design Video Sharing Platform/README_EN.md
@@ -40,7 +40,7 @@ videoSharingPlatform.remove(0);              // Remove the video associated with
 videoSharingPlatform.upload(&quot;789&quot;);          // Since the video associated with videoId 0 was deleted,
                                              // 0 is the smallest available <code>videoId</code>, so return 0.
 videoSharingPlatform.watch(1, 0, 5);         // The video associated with videoId 1 is &quot;456&quot;.
-                                             // The video from minute 0 to min(5, 3 - 1) = 2 is &quot;456&quot;, so return &quot;453&quot;.
+                                             // The video from minute 0 to min(5, 3 - 1) = 2 is &quot;456&quot;, so return &quot;456&quot;.
 videoSharingPlatform.watch(1, 0, 1);         // The video associated with videoId 1 is &quot;456&quot;.
                                              // The video from minute 0 to min(1, 3 - 1) = 1 is &quot;45&quot;, so return &quot;45&quot;.
 videoSharingPlatform.like(1);                // Increase the number of likes on the video associated with videoId 1.

diff --git a/...Minimum Number of Operations to Make All Array Elements Equal to 1/README_EN.md b/...Minimum Number of Operations to Make All Array Elements Equal to 1/README_EN.md
@@ -43,11 +43,6 @@
 	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
 </ul>
 
-<p>&nbsp;</p>
-<p><b>Follow-up:</b></p>
-
-<p>The <code>O(n)</code> time complexity&nbsp;solution works, but could you find an <code>O(1)</code> constant time complexity solution?</p>
-
 ## Solutions
 
 ### Solution 1

diff --git a/solution/2900-2999/2990.Loan Types/README.md b/solution/2900-2999/2990.Loan Types/README.md
@@ -32,7 +32,7 @@ loan_id 是这张表具有唯一值的列。
 
 <pre>
 <b>输入：</b>
-Sessions table:
+Loans table:
 +---------+---------+-----------+
 | loan_id | user_id | loan_type |
 +---------+---------+-----------+

diff --git a/solution/2900-2999/2990.Loan Types/README_EN.md b/solution/2900-2999/2990.Loan Types/README_EN.md
@@ -29,7 +29,7 @@ This table contains loan_id, user_id, and loan_type.
 
 <pre>
 <strong>Input:</strong>
-Sessions table:
+Loans table:
 +---------+---------+-----------+
 | loan_id | user_id | loan_type |
 +---------+---------+-----------+