doocs · yanglbme · Dec 7, 2023 · Dec 7, 2023 · Dec 7, 2023
diff --git a/solution/0500-0599/0557.Reverse Words in a String III/README.md b/solution/0500-0599/0557.Reverse Words in a String III/README.md
@@ -20,8 +20,8 @@
 <p><strong>示例 2:</strong></p>
 
 <pre>
-<strong>输入：</strong> s = "God Ding"
-<strong>输出：</strong>"doG gniD"
+<strong>输入：</strong> s = "Mr Ding"
+<strong>输出：</strong>"rM gniD"
 </pre>
 
 <p>&nbsp;</p>

diff --git a/solution/0500-0599/0557.Reverse Words in a String III/README_EN.md b/solution/0500-0599/0557.Reverse Words in a String III/README_EN.md
@@ -8,12 +8,19 @@
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
-<pre><strong>Input:</strong> s = "Let's take LeetCode contest"
-<strong>Output:</strong> "s'teL ekat edoCteeL tsetnoc"
-</pre><p><strong class="example">Example 2:</strong></p>
-<pre><strong>Input:</strong> s = "God Ding"
-<strong>Output:</strong> "doG gniD"
+
+<pre>
+<strong>Input:</strong> s = &quot;Let&#39;s take LeetCode contest&quot;
+<strong>Output:</strong> &quot;s&#39;teL ekat edoCteeL tsetnoc&quot;
 </pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;Mr Ding&quot;
+<strong>Output:</strong> &quot;rM gniD&quot;
+</pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

diff --git a/solution/1000-1099/1084.Sales Analysis III/README.md b/solution/1000-1099/1084.Sales Analysis III/README.md
@@ -76,10 +76,10 @@ Product table:
 | 1           | S8           |
 +-------------+--------------+
 <strong>解释:</strong>
-id为1的产品仅在2019年春季销售。
-id为2的产品在2019年春季销售，但也在2019年春季之后销售。
-id 3的产品在2019年春季之后销售。
-我们只退回产品1，因为它是2019年春季才销售的产品。</pre>
+id 为 1 的产品仅在 2019 年春季销售。
+id 为 2 的产品在 2019 年春季销售，但也在 2019 年春季之后销售。
+id 为 3 的产品在 2019 年春季之后销售。
+我们只返回 id 为 1 的产品，因为它是 2019 年春季才销售的产品。</pre>
 
 ## 解法
 

diff --git a/solution/1500-1599/1522.Diameter of N-Ary Tree/README.md b/solution/1500-1599/1522.Diameter of N-Ary Tree/README.md
@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定一棵 N 叉树的根节点&nbsp;<code>root</code>&nbsp;，计算这棵树的直径长度。</p>
+<p>给定一棵 <code>N 叉树</code> 的根节点&nbsp;<code>root</code>&nbsp;，计算这棵树的直径长度。</p>
 
 <p>N 叉树的直径指的是树中任意两个节点间路径中<strong> 最长 </strong>路径的长度。这条路径可能经过根节点，也可能不经过根节点。</p>
 

diff --git a/solution/1500-1599/1522.Diameter of N-Ary Tree/README_EN.md b/solution/1500-1599/1522.Diameter of N-Ary Tree/README_EN.md
@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Given a <code>root</code> of an <a href="https://leetcode.com/articles/introduction-to-n-ary-trees/" target="_blank">N-ary tree</a>, you need to compute the length of the diameter of the tree.</p>
+<p>Given a <code>root</code> of an <code>N-ary tree</code>, you need to compute the length of the diameter of the tree.</p>
 
 <p>The diameter of an N-ary tree is the length of the <strong>longest</strong> path between any two nodes in the tree. This path may or may not pass through the root.</p>
 

diff --git a/solution/2400-2499/2431.Maximize Total Tastiness of Purchased Fruits/README.md b/solution/2400-2499/2431.Maximize Total Tastiness of Purchased Fruits/README.md
@@ -26,8 +26,8 @@
 <p><strong>注意:</strong></p>
 
 <ul>
-	<li>每种水果最多只能购买一次。</li>
-	<li>一些水果你最多只能用一次折价券。</li>
+	<li>每个水果最多只能购买一次。</li>
+	<li>一个水果你最多只能用一次折价券。</li>
 </ul>
 
 <p>&nbsp;</p>

diff --git a/...2400-2499/2436.Minimum Split Into Subarrays With GCD Greater Than One/README.md b/...2400-2499/2436.Minimum Split Into Subarrays With GCD Greater Than One/README.md
@@ -8,7 +8,7 @@
 
 <p>给定一个由正整数组成的数组 <code>nums</code>。</p>
 
-<p>将数组拆分为&nbsp;<strong>一个或多个&nbsp;</strong>不相连的子数组，如下所示:</p>
+<p>将数组拆分为&nbsp;<strong>一个或多个&nbsp;</strong>互相不覆盖的子数组，如下所示:</p>
 
 <ul>
 	<li>数组中的每个元素都&nbsp;<strong>只属于一个&nbsp;</strong>子数组，并且</li>

diff --git a/solution/2600-2699/2661.First Completely Painted Row or Column/README.md b/solution/2600-2699/2661.First Completely Painted Row or Column/README.md
@@ -10,7 +10,7 @@
 
 <p>从下标 <code>0</code> 开始遍历 <code>arr</code> 中的每个下标 <code>i</code> ，并将包含整数 <code>arr[i]</code> 的 <code>mat</code> 单元格涂色。</p>
 
-<p>请你找出 <code>arr</code> 中在 <code>mat</code> 的某一行或某一列上都被涂色且下标最小的元素，并返回其下标 <code>i</code> 。</p>
+<p>请你找出 <code>arr</code> 中第一个使得&nbsp;<code>mat</code> 的某一行或某一列都被涂色的元素，并返回其下标 <code>i</code> 。</p>
 
 <p>&nbsp;</p>
 

diff --git a/solution/2700-2799/2715.Timeout Cancellation/README_EN.md b/solution/2700-2799/2715.Timeout Cancellation/README_EN.md
@@ -6,53 +6,66 @@
 
 <p>Given a function <code>fn</code>, an array of&nbsp;arguments&nbsp;<code>args</code>, and a timeout&nbsp;<code>t</code>&nbsp;in milliseconds, return a cancel function <code>cancelFn</code>.</p>
 
-<p>After a delay of&nbsp;<code>t</code>,&nbsp;<code>fn</code>&nbsp;should be called with <code>args</code> passed as parameters <strong>unless</strong> <code>cancelFn</code> was invoked before the delay of <code>t</code> milliseconds elapses, specifically at <code>cancelT</code>&nbsp;ms.&nbsp;In that case,&nbsp;<code>fn</code> should never be called.</p>
+<p>After a delay of <code>cancelT</code>, the returned cancel function <code>cancelFn</code> will be invoked.</p>
+
+<pre>
+setTimeout(cancelFn, cancelT)
+</pre>
+
+<p>Initially, the execution of the function <code>fn</code> should be delayed by <code>t</code> milliseconds.</p>
+
+<p>If, before the delay of <code>t</code> milliseconds, the function <code>cancelFn</code> is invoked, it should cancel the delayed execution of <code>fn</code>. Otherwise, if <code>cancelFn</code> is not invoked within the specified delay <code>t</code>, <code>fn</code> should be executed with the provided <code>args</code> as arguments.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x) =&gt; x * 5, args = [2], t = 20, cancelT = 50
+<strong>Input:</strong> fn = (x) =&gt; x * 5, args = [2], t = 20
 <strong>Output:</strong> [{&quot;time&quot;: 20, &quot;returned&quot;: 10}]
 <strong>Explanation:</strong> 
-const result = []
+const cancelT = 50;
+const result = [];
 
-const fn = (x) =&gt; x * 5
+const fn = (x) =&gt; x * 5;
 
-const start = performance.now()&nbsp;
+const start = performance.now();
 
 const log = (...argsArr) =&gt; {
     const diff = Math.floor(performance.now() - start);
-    result.push({&quot;time&quot;: diff, &quot;returned&quot;: fn(...argsArr)})
+    result.push({&quot;time&quot;: diff, &quot;returned&quot;: fn(...argsArr)});
 }
  &nbsp; &nbsp;&nbsp;
 const cancel = cancellable(log, [2], 20);
 
-const maxT = Math.max(t, 50)
+const maxT = Math.max(t, 50);
 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;
-setTimeout(cancel, cancelT)
+setTimeout(cancel, cancelT);
 
 setTimeout(() =&gt; {
- &nbsp; &nbsp; console.log(result) // [{&quot;time&quot;:20,&quot;returned&quot;:10}]
-}, 65)
+ &nbsp; &nbsp; console.log(result); // [{&quot;time&quot;:20,&quot;returned&quot;:10}]
+}, 65);
 
 The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened after the execution of fn(2) at 20ms.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x) =&gt; x**2, args = [2], t = 100, cancelT = 50 
+<strong>Input:</strong> fn = (x) =&gt; x**2, args = [2], t = 100
 <strong>Output:</strong> []
-<strong>Explanation:</strong> The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened before the execution of fn(2) at 100ms, resulting in fn(2) never being called.
+<strong>Explanation:</strong> 
+const cancelT = 50;
+The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened before the execution of fn(2) at 100ms, resulting in fn(2) never being called.
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30, cancelT = 100
+<strong>Input:</strong> fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30
 <strong>Output:</strong> [{&quot;time&quot;: 30, &quot;returned&quot;: 8}]
-<strong>Explanation: </strong>The cancellation was scheduled to occur after a delay of cancelT (100ms), which happened after the execution of fn(2,4) at 30ms.
+<strong>Explanation: 
+</strong>const cancelT = 100;
+The cancellation was scheduled to occur after a delay of cancelT (100ms), which happened after the execution of fn(2,4) at 30ms.
 </pre>
 
 <p>&nbsp;</p>

diff --git a/...-2999/2948.Make Lexicographically Smallest Array by Swapping Elements/README.md b/...-2999/2948.Make Lexicographically Smallest Array by Swapping Elements/README.md
@@ -12,7 +12,7 @@
 
 <p>返回执行任意次操作后能得到的 <strong>字典序最小的数组</strong><em> </em>。</p>
 
-<p>如果在数组 <code>a</code> 和数组 <code>b</code> 第一个不同的位置上，数组 <code>a</code> 中的对应字符比数组 <code>b</code> 中的对应字符的字典序更小，则认为数组 <code>a</code> 就比数组 <code>b</code> 字典序更小。例如，数组 <code>[2,10,3]</code> 比数组 <code>[10,2,3]</code> 字典序更小，下标 <code>0</code> 处是两个数组第一个不同的位置，且 <code>2 &lt; 10</code> 。</p>
+<p>如果在数组 <code>a</code> 和数组 <code>b</code> 第一个不同的位置上，数组 <code>a</code> 中的对应元素比数组 <code>b</code> 中的对应元素的字典序更小，则认为数组 <code>a</code> 就比数组 <code>b</code> 字典序更小。例如，数组 <code>[2,10,3]</code> 比数组 <code>[10,2,3]</code> 字典序更小，下标 <code>0</code> 处是两个数组第一个不同的位置，且 <code>2 &lt; 10</code> 。</p>
 
 <p>&nbsp;</p>
 

diff --git a/solution/2900-2999/2955.Number of Same-End Substrings/README.md b/solution/2900-2999/2955.Number of Same-End Substrings/README.md
@@ -0,0 +1,90 @@
+# [2955. Number of Same-End Substrings](https://leetcode.cn/problems/number-of-same-end-substrings)
+
+[English Version](/solution/2900-2999/2955.Number%20of%20Same-End%20Substrings/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>You are given a <strong>0-indexed</strong> string <code>s</code>, and a 2D array of integers <code>queries</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code> indicates a substring of <code>s</code> starting from the index <code>l<sub>i</sub></code> and ending at the index <code>r<sub>i</sub></code> (both <strong>inclusive</strong>), i.e. <code>s[l<sub>i</sub>..r<sub>i</sub>]</code>.</p>
+
+<p>Return <em>an array </em><code>ans</code><em> where</em> <code>ans[i]</code> <em>is the number of <strong>same-end</strong> substrings of</em> <code>queries[i]</code>.</p>
+
+<p>A <strong>0-indexed</strong> string <code>t</code> of length <code>n</code> is called <strong>same-end</strong> if it has the same character at both of its ends, i.e., <code>t[0] == t[n - 1]</code>.</p>
+
+<p>A <b>substring</b> is a contiguous non-empty sequence of characters within a string.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcaab&quot;, queries = [[0,0],[1,4],[2,5],[0,5]]
+<strong>Output:</strong> [1,5,5,10]
+<strong>Explanation:</strong> Here is the same-end substrings of each query:
+1<sup>st</sup> query: s[0..0] is &quot;a&quot; which has 1 same-end substring: &quot;<strong><u>a</u></strong>&quot;.
+2<sup>nd</sup> query: s[1..4] is &quot;bcaa&quot; which has 5 same-end substrings: &quot;<strong><u>b</u></strong>caa&quot;, &quot;b<strong><u>c</u></strong>aa&quot;, &quot;bc<strong><u>a</u></strong>a&quot;, &quot;bca<strong><u>a</u></strong>&quot;, &quot;bc<strong><u>aa</u></strong>&quot;.
+3<sup>rd</sup> query: s[2..5] is &quot;caab&quot; which has 5 same-end substrings: &quot;<strong><u>c</u></strong>aab&quot;, &quot;c<strong><u>a</u></strong>ab&quot;, &quot;ca<strong><u>a</u></strong>b&quot;, &quot;caa<strong><u>b</u></strong>&quot;, &quot;c<strong><u>aa</u></strong>b&quot;.
+4<sup>th</sup> query: s[0..5] is &quot;abcaab&quot; which has 10 same-end substrings: &quot;<strong><u>a</u></strong>bcaab&quot;, &quot;a<strong><u>b</u></strong>caab&quot;, &quot;ab<strong><u>c</u></strong>aab&quot;, &quot;abc<strong><u>a</u></strong>ab&quot;, &quot;abca<strong><u>a</u></strong>b&quot;, &quot;abcaa<strong><u>b</u></strong>&quot;, &quot;abc<strong><u>aa</u></strong>b&quot;, &quot;<strong><u>abca</u></strong>ab&quot;, &quot;<strong><u>abcaa</u></strong>b&quot;, &quot;a<strong><u>bcaab</u></strong>&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcd&quot;, queries = [[0,3]]
+<strong>Output:</strong> [4]
+<strong>Explanation:</strong> The only query is s[0..3] which is &quot;abcd&quot;. It has 4 same-end substrings: &quot;<strong><u>a</u></strong>bcd&quot;, &quot;a<strong><u>b</u></strong>cd&quot;, &quot;ab<strong><u>c</u></strong>d&quot;, &quot;abc<strong><u>d</u></strong>&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>s</code> consists only of lowercase English letters.</li>
+	<li><code>1 &lt;= queries.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>queries[i].length == 2</code></li>
+	<li><code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code></li>
+	<li><code>0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt; s.length</code></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 --git a/solution/2900-2999/2955.Number of Same-End Substrings/README_EN.md b/solution/2900-2999/2955.Number of Same-End Substrings/README_EN.md
@@ -0,0 +1,82 @@
+# [2955. Number of Same-End Substrings](https://leetcode.com/problems/number-of-same-end-substrings)
+
+[中文文档](/solution/2900-2999/2955.Number%20of%20Same-End%20Substrings/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> string <code>s</code>, and a 2D array of integers <code>queries</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code> indicates a substring of <code>s</code> starting from the index <code>l<sub>i</sub></code> and ending at the index <code>r<sub>i</sub></code> (both <strong>inclusive</strong>), i.e. <code>s[l<sub>i</sub>..r<sub>i</sub>]</code>.</p>
+
+<p>Return <em>an array </em><code>ans</code><em> where</em> <code>ans[i]</code> <em>is the number of <strong>same-end</strong> substrings of</em> <code>queries[i]</code>.</p>
+
+<p>A <strong>0-indexed</strong> string <code>t</code> of length <code>n</code> is called <strong>same-end</strong> if it has the same character at both of its ends, i.e., <code>t[0] == t[n - 1]</code>.</p>
+
+<p>A <b>substring</b> is a contiguous non-empty sequence of characters within a string.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcaab&quot;, queries = [[0,0],[1,4],[2,5],[0,5]]
+<strong>Output:</strong> [1,5,5,10]
+<strong>Explanation:</strong> Here is the same-end substrings of each query:
+1<sup>st</sup> query: s[0..0] is &quot;a&quot; which has 1 same-end substring: &quot;<strong><u>a</u></strong>&quot;.
+2<sup>nd</sup> query: s[1..4] is &quot;bcaa&quot; which has 5 same-end substrings: &quot;<strong><u>b</u></strong>caa&quot;, &quot;b<strong><u>c</u></strong>aa&quot;, &quot;bc<strong><u>a</u></strong>a&quot;, &quot;bca<strong><u>a</u></strong>&quot;, &quot;bc<strong><u>aa</u></strong>&quot;.
+3<sup>rd</sup> query: s[2..5] is &quot;caab&quot; which has 5 same-end substrings: &quot;<strong><u>c</u></strong>aab&quot;, &quot;c<strong><u>a</u></strong>ab&quot;, &quot;ca<strong><u>a</u></strong>b&quot;, &quot;caa<strong><u>b</u></strong>&quot;, &quot;c<strong><u>aa</u></strong>b&quot;.
+4<sup>th</sup> query: s[0..5] is &quot;abcaab&quot; which has 10 same-end substrings: &quot;<strong><u>a</u></strong>bcaab&quot;, &quot;a<strong><u>b</u></strong>caab&quot;, &quot;ab<strong><u>c</u></strong>aab&quot;, &quot;abc<strong><u>a</u></strong>ab&quot;, &quot;abca<strong><u>a</u></strong>b&quot;, &quot;abcaa<strong><u>b</u></strong>&quot;, &quot;abc<strong><u>aa</u></strong>b&quot;, &quot;<strong><u>abca</u></strong>ab&quot;, &quot;<strong><u>abcaa</u></strong>b&quot;, &quot;a<strong><u>bcaab</u></strong>&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcd&quot;, queries = [[0,3]]
+<strong>Output:</strong> [4]
+<strong>Explanation:</strong> The only query is s[0..3] which is &quot;abcd&quot;. It has 4 same-end substrings: &quot;<strong><u>a</u></strong>bcd&quot;, &quot;a<strong><u>b</u></strong>cd&quot;, &quot;ab<strong><u>c</u></strong>d&quot;, &quot;abc<strong><u>d</u></strong>&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>s</code> consists only of lowercase English letters.</li>
+	<li><code>1 &lt;= queries.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>queries[i].length == 2</code></li>
+	<li><code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code></li>
+	<li><code>0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt; s.length</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->