feat: update lc problems (#4123)

Author: yanglbme

solution/0200-0299/0262.Trips and Users/README.md

@@ -61,7 +61,7 @@ banned 是一个表示用户是否被禁止的枚举类型，枚举成员为 (
 
 <p><strong>取消率</strong> 的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)。</p>
 
-<p>编写解决方案找出&nbsp;<code>"2013-10-01"</code><strong>&nbsp;</strong>至&nbsp;<code>"2013-10-03"</code><strong>&nbsp;</strong>期间非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的取消率。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。其中取消率 <code>Cancellation Rate</code> 需要四舍五入保留 <strong>两位小数</strong> 。</p>
+<p>编写解决方案找出&nbsp;<code>"2013-10-01"</code><strong>&nbsp;</strong>至&nbsp;<code>"2013-10-03"</code><strong>&nbsp;</strong>期间有&nbsp;<strong>至少&nbsp;</strong>一次行程的非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的 <strong>取消率</strong>。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。其中取消率 <code>Cancellation Rate</code> 需要四舍五入保留 <strong>两位小数</strong> 。</p>
 
 <p>返回结果表中的数据 <strong>无顺序要求</strong> 。</p>
 

solution/1300-1399/1363.Largest Multiple of Three/README.md

@@ -7,7 +7,9 @@ source: 第 177 场周赛 Q4
 tags:
     - 贪心
     - 数组
+    - 数学
     - 动态规划
+    - 排序
 ---
 
 <!-- problem:start -->

solution/1300-1399/1363.Largest Multiple of Three/README_EN.md

@@ -7,7 +7,9 @@ source: Weekly Contest 177 Q4
 tags:
     - Greedy
     - Array
+    - Math
     - Dynamic Programming
+    - Sorting
 ---
 
 <!-- problem:start -->

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/README.md

@@ -7,7 +7,7 @@ source: 第 178 场周赛 Q1
 tags:
     - 数组
     - 哈希表
-    - 计数
+    - 计数排序
     - 排序
 ---
 

solution/1300-1399/1365.How Many Numbers Are Smaller Than the Current Number/README_EN.md

@@ -7,7 +7,7 @@ source: Weekly Contest 178 Q1
 tags:
     - Array
     - Hash Table
-    - Counting
+    - Counting Sort
     - Sorting
 ---
 

solution/1400-1499/1434.Number of Ways to Wear Different Hats to Each Other/README.md

@@ -57,13 +57,6 @@ tags:
 (1,2,3,4) 4 个帽子的排列方案数为 24 。
 </pre>
 
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>hats = [[1,2,3],[2,3,5,6],[1,3,7,9],[1,8,9],[2,5,7]]
-<strong>输出：</strong>111
-</pre>
-
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>

solution/1400-1499/1434.Number of Ways to Wear Different Hats to Each Other/README_EN.md

@@ -25,7 +25,7 @@ tags:
 
 <p>Given a 2D integer array <code>hats</code>, where <code>hats[i]</code> is a list of all hats preferred by the <code>i<sup>th</sup></code> person.</p>
 
-<p>Return <em>the number of ways that the <code>n</code> people wear different hats to each other</em>.</p>
+<p>Return the number of ways that <code>n</code> people can wear <strong>different</strong> hats from each other.</p>
 
 <p>Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>
 

solution/1900-1999/1968.Array With Elements Not Equal to Average of Neighbors/README_EN.md

@@ -47,7 +47,7 @@ When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.
 When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
 When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
 When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.
-</pre>
+Note that the original array [6,2,0,9,7] also satisfies the conditions.</pre>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/1900-1999/1974.Minimum Time to Type Word Using Special Typewriter/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>Input:</strong> word = &quot;abc&quot;
 <strong>Output:</strong> 5
-<strong>Explanation:
+<strong>Explanation: 
 </strong>The characters are printed as follows:
 - Type the character &#39;a&#39; in 1 second since the pointer is initially on &#39;a&#39;.
 - Move the pointer clockwise to &#39;b&#39; in 1 second.

solution/2100-2199/2138.Divide a String Into Groups of Size k/README_EN.md

@@ -22,7 +22,7 @@ tags:
 <p>A string <code>s</code> can be partitioned into groups of size <code>k</code> using the following procedure:</p>
 
 <ul>
-	<li>The first group consists of the first <code>k</code> characters of the string, the second group consists of the next <code>k</code> characters of the string, and so on. Each character can be a part of <strong>exactly one</strong> group.</li>
+	<li>The first group consists of the first <code>k</code> characters of the string, the second group consists of the next <code>k</code> characters of the string, and so on. Each element can be a part of <strong>exactly one</strong> group.</li>
 	<li>For the last group, if the string <strong>does not</strong> have <code>k</code> characters remaining, a character <code>fill</code> is used to complete the group.</li>
 </ul>
 

solution/2200-2299/2276.Count Integers in Intervals/README_EN.md

@@ -48,7 +48,7 @@ tags:
 [null, null, null, 6, null, 8]
 
 <strong>Explanation</strong>
-CountIntervals countIntervals = new CountIntervals(); // initialize the object with an empty set of intervals.
+CountIntervals countIntervals = new CountIntervals(); // initialize the object with an empty set of intervals. 
 countIntervals.add(2, 3);  // add [2, 3] to the set of intervals.
 countIntervals.add(7, 10); // add [7, 10] to the set of intervals.
 countIntervals.count();    // return 6

solution/2200-2299/2287.Rearrange Characters to Make Target String/README.md

@@ -44,7 +44,7 @@ tags:
 <strong>输入：</strong>s = "abcba", target = "abc"
 <strong>输出：</strong>1
 <strong>解释：</strong>
-选取下标为 0 、1 和 2 的字符，可以形成 "abc" 的 1 个副本。
+选取下标为 0 、1 和 2 的字符，可以形成 "abc" 的 1 个副本。 
 可以形成最多 1 个 "abc" 的副本，所以返回 1 。
 注意，尽管下标 3 和 4 分别有额外的 'a' 和 'b' ，但不能重用下标 2 处的 'c' ，所以无法形成 "abc" 的第 2 个副本。
 </pre>

solution/2200-2299/2296.Design a Text Editor/README.md

@@ -62,7 +62,7 @@ textEditor.deleteText(4); // 返回 4
                           // 删除了 4 个字符。
 textEditor.addText("practice"); // 当前文本为 "leetpractice|" 。
 textEditor.cursorRight(3); // 返回 "etpractice"
-                           // 当前文本为 "leetpractice|".
+                           // 当前文本为 "leetpractice|". 
                            // 光标无法移动到文本以外，所以无法移动。
                            // "etpractice" 是光标左边的 10 个字符。
 textEditor.cursorLeft(8); // 返回 "leet"

solution/2200-2299/2296.Design a Text Editor/README_EN.md

@@ -57,11 +57,11 @@ tags:
 TextEditor textEditor = new TextEditor(); // The current text is "|". (The '|' character represents the cursor)
 textEditor.addText("leetcode"); // The current text is "leetcode|".
 textEditor.deleteText(4); // return 4
-                          // The current text is "leet|".
+                          // The current text is "leet|". 
                           // 4 characters were deleted.
-textEditor.addText("practice"); // The current text is "leetpractice|".
+textEditor.addText("practice"); // The current text is "leetpractice|". 
 textEditor.cursorRight(3); // return "etpractice"
-                           // The current text is "leetpractice|".
+                           // The current text is "leetpractice|". 
                            // The cursor cannot be moved beyond the actual text and thus did not move.
                            // "etpractice" is the last 10 characters to the left of the cursor.
 textEditor.cursorLeft(8); // return "leet"
@@ -72,7 +72,7 @@ textEditor.deleteText(10); // return 4
                            // Only 4 characters were deleted.
 textEditor.cursorLeft(2); // return ""
                           // The current text is "|practice".
-                          // The cursor cannot be moved beyond the actual text and thus did not move.
+                          // The cursor cannot be moved beyond the actual text and thus did not move. 
                           // "" is the last min(10, 0) = 0 characters to the left of the cursor.
 textEditor.cursorRight(6); // return "practi"
                            // The current text is "practi|ce".

solution/2300-2399/2319.Check if Matrix Is X-Matrix/README_EN.md

@@ -34,7 +34,7 @@ tags:
 <pre>
 <strong>Input:</strong> grid = [[2,0,0,1],[0,3,1,0],[0,5,2,0],[4,0,0,2]]
 <strong>Output:</strong> true
-<strong>Explanation:</strong> Refer to the diagram above.
+<strong>Explanation:</strong> Refer to the diagram above. 
 An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0.
 Thus, grid is an X-Matrix.
 </pre>

solution/2300-2399/2320.Count Number of Ways to Place Houses/README_EN.md

@@ -30,7 +30,7 @@ tags:
 <pre>
 <strong>Input:</strong> n = 1
 <strong>Output:</strong> 4
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 Possible arrangements:
 1. All plots are empty.
 2. A house is placed on one side of the street.

solution/2500-2599/2562.Find the Array Concatenation Value/README.md

@@ -31,8 +31,8 @@ tags:
 <p><code>nums</code>&nbsp;的 <strong>串联值</strong>&nbsp;最初等于 <code>0</code> 。执行下述操作直到&nbsp;<code>nums</code>&nbsp;变为空：</p>
 
 <ul>
-	<li>如果&nbsp;<code>nums</code>&nbsp;中存在不止一个数字，分别选中 <code>nums</code> 中的第一个元素和最后一个元素，将二者串联得到的值加到&nbsp;<code>nums</code>&nbsp;的 <strong>串联值</strong> 上，然后从&nbsp;<code>nums</code>&nbsp;中删除第一个和最后一个元素。</li>
-	<li>如果仅存在一个元素，则将该元素的值加到&nbsp;<code>nums</code> 的串联值上，然后删除这个元素。</li>
+	<li>如果&nbsp;<code>nums</code>&nbsp;的长度大于 1，分别选中 <code>nums</code> 中的第一个元素和最后一个元素，将二者串联得到的值加到&nbsp;<code>nums</code>&nbsp;的 <strong>串联值</strong> 上，然后从&nbsp;<code>nums</code>&nbsp;中删除第一个和最后一个元素。例如，如果&nbsp;<code>nums</code> 是 <code>[1, 2, 4, 5, 6]</code>，将 16 添加到串联值。</li>
+	<li>如果&nbsp;<code>nums</code>&nbsp;中仅存在一个元素，则将该元素的值加到&nbsp;<code>nums</code> 的串联值上，然后删除这个元素。</li>
 </ul>
 
 <p>返回执行完所有操作后<em>&nbsp;</em><code>nums</code> 的串联值。</p>

solution/2500-2599/2562.Find the Array Concatenation Value/README_EN.md

@@ -31,11 +31,11 @@ tags:
 <p>The <strong>concatenation value</strong> of <code>nums</code> is initially equal to <code>0</code>. Perform this operation until <code>nums</code> becomes empty:</p>
 
 <ul>
-	<li>If there exists more than one number in <code>nums</code>, pick the first element and last element in <code>nums</code> respectively and add the value of their concatenation to the <strong>concatenation value</strong> of <code>nums</code>, then delete the first and last element from <code>nums</code>.</li>
-	<li>If one element exists, add its value to the <strong>concatenation value</strong> of <code>nums</code>, then delete it.</li>
+	<li>If <code>nums</code> has a size greater than one, add the value of the concatenation of the first and the last element to the <strong>concatenation value</strong> of <code>nums</code>, and remove those two elements from <code>nums</code>. For example, if the <code>nums</code> was <code>[1, 2, 4, 5, 6]</code>, add 16 to the <code>concatenation value</code>.</li>
+	<li>If only one element exists in <code>nums</code>, add its value to the <strong>concatenation value</strong> of <code>nums</code>, then remove it.</li>
 </ul>
 
-<p>Return<em> the concatenation value of the <code>nums</code></em>.</p>
+<p>Return<em> the concatenation value of <code>nums</code></em>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/2500-2599/2574.Left and Right Sum Differences/README.md

@@ -19,35 +19,32 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，请你找出一个下标从 <strong>0</strong> 开始的整数数组 <code>answer</code> ，其中：</p>
+<p>给你一个下标从 <strong>0</strong> 开始的长度为&nbsp;<code>n</code>&nbsp;的整数数组 <code>nums</code>。</p>
 
-<ul>
-	<li><code>answer.length == nums.length</code></li>
-	<li><code>answer[i] = |leftSum[i] - rightSum[i]|</code></li>
-</ul>
-
-<p>其中：</p>
+<p>定义两个数组&nbsp;<code>leftSum</code>&nbsp;和&nbsp;<code>rightSum</code>，其中：</p>
 
 <ul>
 	<li><code>leftSum[i]</code> 是数组 <code>nums</code> 中下标 <code>i</code> 左侧元素之和。如果不存在对应的元素，<code>leftSum[i] = 0</code> 。</li>
 	<li><code>rightSum[i]</code> 是数组 <code>nums</code> 中下标 <code>i</code> 右侧元素之和。如果不存在对应的元素，<code>rightSum[i] = 0</code> 。</li>
 </ul>
 
-<p>返回数组 <code>answer</code> 。</p>
+<p>返回长度为&nbsp;<code>n</code> 数组 <code>answer</code>，其中 <code>answer[i] = |leftSum[i] - rightSum[i]|</code>。</p>
 
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入：</strong>nums = [10,4,8,3]
+<pre>
+<strong>输入：</strong>nums = [10,4,8,3]
 <strong>输出：</strong>[15,1,11,22]
 <strong>解释：</strong>数组 leftSum 为 [0,10,14,22] 且数组 rightSum 为 [15,11,3,0] 。
 数组 answer 为 [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22] 。
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
-<pre><strong>输入：</strong>nums = [1]
+<pre>
+<strong>输入：</strong>nums = [1]
 <strong>输出：</strong>[0]
 <strong>解释：</strong>数组 leftSum 为 [0] 且数组 rightSum 为 [0] 。
 数组 answer 为 [|0 - 0|] = [0] 。

solution/2500-2599/2574.Left and Right Sum Differences/README_EN.md

@@ -19,21 +19,16 @@ tags:
 
 <!-- description:start -->
 
-<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, find a <strong>0-indexed </strong>integer array <code>answer</code> where:</p>
+<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code>.</p>
 
-<ul>
-    <li><code>answer.length == nums.length</code>.</li>
-    <li><code>answer[i] = |leftSum[i] - rightSum[i]|</code>.</li>
-</ul>
-
-<p>Where:</p>
+<p>Define two arrays <code>leftSum</code> and <code>rightSum</code> where:</p>
 
 <ul>
-    <li><code>leftSum[i]</code> is the sum of elements to the left of the index <code>i</code> in the array <code>nums</code>. If there is no such element, <code>leftSum[i] = 0</code>.</li>
-    <li><code>rightSum[i]</code> is the sum of elements to the right of the index <code>i</code> in the array <code>nums</code>. If there is no such element, <code>rightSum[i] = 0</code>.</li>
+	<li><code>leftSum[i]</code> is the sum of elements to the left of the index <code>i</code> in the array <code>nums</code>. If there is no such element, <code>leftSum[i] = 0</code>.</li>
+	<li><code>rightSum[i]</code> is the sum of elements to the right of the index <code>i</code> in the array <code>nums</code>. If there is no such element, <code>rightSum[i] = 0</code>.</li>
 </ul>
 
-<p>Return <em>the array</em> <code>answer</code>.</p>
+<p>Return an integer array <code>answer</code> of size <code>n</code> where <code>answer[i] = |leftSum[i] - rightSum[i]|</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/3100-3199/3147.Taking Maximum Energy From the Mystic Dungeon/README_EN.md

@@ -27,6 +27,8 @@ tags:
 
 <p>You are given an array <code>energy</code> and an integer <code>k</code>. Return the <strong>maximum</strong> possible energy you can gain.</p>
 
+<p><strong>Note</strong> that when you are reach a magician, you <em>must</em> take energy from them, whether it is negative or positive energy.</p>
+
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>