chore: update lc problems (doocs#1950)

Author: yanglbme

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/README.md

@@ -17,7 +17,7 @@
 <p><strong>示例 1：</strong></p>
 
 <pre>
-<b>输入：</b>beans = [4,<u><strong>1</strong></u>,6,5]
+<b>输入：</b>beans = [4,1,6,5]
 <b>输出：</b>4
 <b>解释：</b>
 - 我们从有 1 个魔法豆的袋子中拿出 1 颗魔法豆。
@@ -33,7 +33,7 @@
 <p><strong>示例 2：</strong></p>
 
 <pre>
-<b>输入：</b>beans = [<u><strong>2</strong></u>,10,<u><strong>3</strong></u>,<u><strong>2</strong></u>]
+<b>输入：</b>beans = [2,10,3,2]
 <b>输出：</b>7
 <strong>解释：</strong>
 - 我们从有 2 个魔法豆的其中一个袋子中拿出 2 个魔法豆。

solution/2900-2999/2921.Maximum Profitable Triplets With Increasing Prices II/README.md

@@ -1,52 +1,54 @@
-# [2921. Maximum Profitable Triplets With Increasing Prices II](https://leetcode.cn/problems/maximum-profitable-triplets-with-increasing-prices-ii)
+# [2921. 具有递增价格的最大利润三元组 II](https://leetcode.cn/problems/maximum-profitable-triplets-with-increasing-prices-ii)
 
 [English Version](/solution/2900-2999/2921.Maximum%20Profitable%20Triplets%20With%20Increasing%20Prices%20II/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>Given the <strong>0-indexed</strong> arrays <code>prices</code> and <code>profits</code> of length <code>n</code>. There are <code>n</code> items in an store where the <code>i<sup>th</sup></code> item has a price of <code>prices[i]</code> and a profit of <code>profits[i]</code>.</p>
+<p>给定长度为 <code>n</code>&nbsp; 的数组&nbsp;<code>prices</code>&nbsp;和&nbsp;<code>profits</code>&nbsp;（<strong>下标从 0 开始</strong>）。一个商店有&nbsp;<code>n</code>&nbsp;个商品，第&nbsp;<code>i</code>&nbsp;个商品的价格为&nbsp;<code>prices[i]</code>，利润为&nbsp;<code>profits[i]</code>。</p>
 
-<p>We have to pick three items with the following condition:</p>
+<p>需要选择三个商品，满足以下条件：</p>
 
 <ul>
-	<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code> where <code>i &lt; j &lt; k</code>.</li>
+	<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code>，其中&nbsp;<code>i &lt; j &lt; k</code>。</li>
 </ul>
 
-<p>If we pick items with indices <code>i</code>, <code>j</code> and <code>k</code> satisfying the above condition, the profit would be <code>profits[i] + profits[j] + profits[k]</code>.</p>
+<p>如果选择的商品&nbsp;<code>i</code>, <code>j</code>&nbsp;和&nbsp;<code>k</code>&nbsp;满足以下条件，那么利润将等于&nbsp;<code>profits[i] + profits[j] + profits[k]</code>。</p>
 
-<p>Return<em> the <strong>maximum profit</strong> we can get, and </em><code>-1</code><em> if it&#39;s not possible to pick three items with the given condition.</em></p>
+<p>返回能够获得的<em>&nbsp;<strong>最大利润</strong>，如果不可能满足给定条件，</em>返回<em>&nbsp;</em><code>-1</code><em>。</em></p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>Input:</strong> prices = [10,2,3,4], profits = [100,2,7,10]
-<strong>Output:</strong> 19
-<strong>Explanation:</strong> We can&#39;t pick the item with index i=0 since there are no indices j and k such that the condition holds.
-So the only triplet we can pick, are the items with indices 1, 2 and 3 and it&#39;s a valid pick since prices[1] &lt; prices[2] &lt; prices[3].
-The answer would be sum of their profits which is 2 + 7 + 10 = 19.</pre>
+<b>输入：</b>prices = [10,2,3,4], profits = [100,2,7,10]
+<b>输出：</b>19
+<b>解释：</b>不能选择下标 i=0 的商品，因为没有下标 j 和 k 的商品满足条件。
+只能选择下标为 1、2、3 的三个商品，这是有效的选择，因为 prices[1] &lt; prices[2] &lt; prices[3]。
+答案是它们的利润之和，即 2 + 7 + 10 = 19。</pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>Input:</strong> prices = [1,2,3,4,5], profits = [1,5,3,4,6]
-<strong>Output:</strong> 15
-<strong>Explanation:</strong> We can select any triplet of items since for each triplet of indices i, j and k such that i &lt; j &lt; k, the condition holds.
-Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4.
-The answer would be sum of their profits which is 5 + 4 + 6 = 15.</pre>
+<b>输入：</b>prices = [1,2,3,4,5], profits = [1,5,3,4,6]
+<b>输出：</b>15
+<b>解释：</b>可以选择任意三个商品，因为对于每组满足下标为 i &lt; j &lt; k 的三个商品，条件都成立。
+因此，能得到的最大利润就是利润和最大的三个商品，即下标为 1，3 和 4 的商品。
+答案就是它们的利润之和，即 5 + 4 + 6 = 15。</pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>示例 3：</strong></p>
 
 <pre>
-<strong>Input:</strong> prices = [4,3,2,1], profits = [33,20,19,87]
-<strong>Output:</strong> -1
-<strong>Explanation:</strong> We can&#39;t select any triplet of indices such that the condition holds, so we return -1.
+<b>输入：</b>prices = [4,3,2,1], profits = [33,20,19,87]
+<b>输出：</b>-1
+<b>解释：</b>找不到任何可以满足条件的三个商品，所以返回 -1.
 </pre>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><b>提示：</b></p>
 
 <ul>
 	<li><code>3 &lt;= prices.length == profits.length &lt;= 50000</code></li>

solution/2900-2999/2922.Market Analysis III/README.md

@@ -1,4 +1,4 @@
-# [2922. Market Analysis III](https://leetcode.cn/problems/market-analysis-iii)
+# [2922. 市场分析 III](https://leetcode.cn/problems/market-analysis-iii)
 
 [English Version](/solution/2900-2999/2922.Market%20Analysis%20III/README_EN.md)
 
@@ -16,8 +16,8 @@
 | join_date      | date    |
 | favorite_brand | varchar |
 +----------------+---------+
-seller_id is the primary key for this table.
-This table contains seller id, join date, and favorite brand of sellers.
+seller_id 是该表的主键。
+该表包含销售者的 ID, 加入日期以及最喜欢的品牌。
 </pre>
 
 <p>Table: <code>Items</code></p>
@@ -29,8 +29,8 @@ This table contains seller id, join date, and favorite brand of sellers.
 | item_id       | int     |
 | item_brand    | varchar |
 +---------------+---------+
-item_id is the primary key for this table.
-This table contains item id and item brand.</pre>
+item_id 是该表的主键。
+该表包含商品 ID 和商品品牌。</pre>
 
 <p>Table: <code>Orders</code></p>
 
@@ -43,22 +43,23 @@ This table contains item id and item brand.</pre>
 | item_id       | int     |
 | seller_id     | int     |
 +---------------+---------+
-order_id is the primary key for this table.
-item_id is a foreign key to the Items table.
-seller_id is a foreign key to the Users table.
-This table contains order id, order date, item id and seller id.</pre>
+order_id 是该表的主键。
+item_id 是指向 Items 表的外键。
+seller_id 是指向 Users 表的外键。
+该表包含订单 ID、下单日期、商品 ID 和卖家 ID。</pre>
 
-<p>Write a solution to find the <strong>top seller</strong> who has sold the highest number of<strong> unique</strong> items with a <strong>different</strong> brand than their favorite brand. If there are multiple sellers with the same highest count, return all of them.</p>
+<p>编写一个解决方案，找到销售最多与其最喜欢的品牌 <strong>不同</strong> 的 <strong>独特</strong> 商品的&nbsp;<strong>顶级卖家</strong>。如果有多个卖家销售同样数量的商品，则返回所有这些卖家。&nbsp;</p>
 
-<p>Return <em>the result table ordered by</em> <code>seller_id</code> <em>in <strong>ascending</strong> order.</em></p>
+<p>返回按&nbsp;<code>seller_id</code><em>&nbsp;<strong>升序排序</strong>&nbsp;的结果表。</em></p>
 
-<p>The result format is in the following example.</p>
+<p>结果格式如下示例所示。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><b>示例 1:</b></p>
 
 <pre>
-<strong>Input:</strong> 
+<b>输入：</b>
 Users table:
 +-----------+------------+----------------+
 | seller_id | join_date  | favorite_brand |
@@ -86,17 +87,18 @@ Items table:
 | 3       | LG         |
 | 4       | HP         |
 +---------+------------+
-<strong>Output:</strong> 
+<b>输出：</b>
 +-----------+-----------+
 | seller_id | num_items |
 +-----------+-----------+
 | 2         | 1         |
 | 3         | 1         |
 +-----------+-----------+
-<strong>Explanation:</strong> 
-- The user with seller_id 2 has sold three items, but only two of them are not marked as a favorite. We will include a unique count of 1 because both of these items are identical.
-- The user with seller_id 3 has sold two items, but only one of them is not marked as a favorite. We will include just that non-favorite item in our count.
-Since seller_ids 2 and 3 have the same count of one item each, they both will be displayed in the output.</pre>
+<b>解释：</b>
+- 卖家 ID 为 2 的用户销售了三件商品，但只有两件不被标记为最喜欢的商品。我们将只列入独特计数为 1，因为这两件商品都是相同的。 
+- 卖家 ID 为 3 的用户销售了两件商品，但只有一件不被标记为最喜欢的商品。我们将只包括那个不被标记为最喜欢的商品列入计数中。
+因为卖家 ID 为 2 和 3 的卖家都有一件商品列入计数，所以他们都将显示在输出中。
+</pre>
 
 ## 解法
 

solution/2900-2999/2923.Find Champion I/README.md

@@ -44,9 +44,10 @@ grid[1][2] == 1 表示 1 队比 2 队强。
 	<li><code>n == grid.length</code></li>
 	<li><code>n == grid[i].length</code></li>
 	<li><code>2 &lt;= n &lt;= 100</code></li>
-	<li><code>grid[i][j]</code> 的值为 <code>0</code> 或 <code>1</code></li>
+	<li><code>grid[i][j]</code> 的值为 <code>0</code> 或 <code>1</code><meta charset="UTF-8" /></li>
+	<li>对于所有&nbsp;<code>i</code>，<code> grid[i][i]</code>&nbsp;等于&nbsp;<code>0.</code></li>
 	<li>对于满足&nbsp;<code>i != j</code> 的所有 <code>i, j</code> ，<code>grid[i][j] != grid[j][i]</code> 均成立</li>
-	<li>生成的输出满足：如果 <code>a</code> 队比 <code>b</code> 队强，<code>b</code> 队比 <code>c</code> 队强，那么 <code>a</code> 队比 <code>c</code> 队强</li>
+	<li>生成的输入满足：如果 <code>a</code> 队比 <code>b</code> 队强，<code>b</code> 队比 <code>c</code> 队强，那么 <code>a</code> 队比 <code>c</code> 队强</li>
 </ul>
 
 ## 解法

solution/2900-2999/2923.Find Champion I/README_EN.md

@@ -41,6 +41,7 @@ So team 1 will be the champion.
 	<li><code>n == grid[i].length</code></li>
 	<li><code>2 &lt;= n &lt;= 100</code></li>
 	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
+	<li>For all <code>i grid[i][i]</code> is <code>0.</code></li>
 	<li>For all <code>i, j</code> that <code>i != j</code>, <code>grid[i][j] != grid[j][i]</code>.</li>
 	<li>The input is generated such that if team <code>a</code> is stronger than team <code>b</code> and team <code>b</code> is stronger than team <code>c</code>, then team <code>a</code> is stronger than team <code>c</code>.</li>
 </ul>

solution/2900-2999/2927.Distribute Candies Among Children III/README.md

@@ -1,34 +1,36 @@
-# [2927. Distribute Candies Among Children III](https://leetcode.cn/problems/distribute-candies-among-children-iii)
+# [2927. 将糖果分发给孩子 III](https://leetcode.cn/problems/distribute-candies-among-children-iii)
 
 [English Version](/solution/2900-2999/2927.Distribute%20Candies%20Among%20Children%20III/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>You are given two positive integers <code>n</code> and <code>limit</code>.</p>
+<p>你被给定两个正整数 <code>n</code> 和 <code>limit</code>。</p>
 
-<p>Return <em>the <strong>total number</strong> of ways to distribute </em><code>n</code> <em>candies among </em><code>3</code><em> children such that no child gets more than </em><code>limit</code><em> candies.</em></p>
+<p>返回 <em>在每个孩子得到不超过&nbsp;</em><code>limit</code><em> 个糖果的情况下，将</em> <code>n</code> <em>个糖果分发给</em>&nbsp;<code>3</code> <em>个孩子的&nbsp;<strong>总方法数</strong>。</em></p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><b>示例 1:</b></p>
 
 <pre>
-<strong>Input:</strong> n = 5, limit = 2
-<strong>Output:</strong> 3
-<strong>Explanation:</strong> There are 3 ways to distribute 5 candies such that no child gets more than 2 candies: (1, 2, 2), (2, 1, 2) and (2, 2, 1).
+<b>输入：</b>n = 5, limit = 2
+<b>输出：</b>3
+<b>解释：</b>有 3 种方式将 5 个糖果分发给 3 个孩子，使得每个孩子得到不超过 2 个糖果：(1, 2, 2), (2, 1, 2) 和 (2, 2, 1)。
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><b>示例 2:</b></p>
 
 <pre>
-<strong>Input:</strong> n = 3, limit = 3
-<strong>Output:</strong> 10
-<strong>Explanation:</strong> There are 10 ways to distribute 3 candies such that no child gets more than 3 candies: (0, 0, 3), (0, 1, 2), (0, 2, 1), (0, 3, 0), (1, 0, 2), (1, 1, 1), (1, 2, 0), (2, 0, 1), (2, 1, 0) and (3, 0, 0).
+<b>输入：</b>n = 3, limit = 3
+<b>输出：</b>10
+<b>解释：</b>有 10 种方式将 3 个糖果分发给 3 个孩子，使得每个孩子得到不超过 3 个糖果：(0, 0, 3), (0, 1, 2), (0, 2, 1), (0, 3, 0), (1, 0, 2), (1, 1, 1), (1, 2, 0), (2, 0, 1), (2, 1, 0) 和 (3, 0, 0)。
 </pre>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><b>提示：</b></p>
 
 <ul>
 	<li><code>1 &lt;= n &lt;= 10<sup>8</sup></code></li>

solution/DATABASE_README.md

@@ -246,7 +246,7 @@
 | 2853 | [最高薪水差异](/solution/2800-2899/2853.Highest%20Salaries%20Difference/README.md)                                                                           | `数据库` | 简单 | 🔒   |
 | 2854 | [滚动平均步数](/solution/2800-2899/2854.Rolling%20Average%20Steps/README.md)                                                                                 | `数据库` | 中等 | 🔒   |
 | 2893 | [计算每个区间内的订单](/solution/2800-2899/2893.Calculate%20Orders%20Within%20Each%20Interval/README.md)                                                     | `数据库` | 中等 | 🔒   |
-| 2922 | [Market Analysis III](/solution/2900-2999/2922.Market%20Analysis%20III/README.md)                                                                            |          | 中等 | 🔒   |
+| 2922 | [市场分析 III](/solution/2900-2999/2922.Market%20Analysis%20III/README.md)                                                                                   |          | 中等 | 🔒   |
 
 ## 版权
 

solution/JAVASCRIPT_README.md

@@ -24,7 +24,7 @@
 | 2629 | [复合函数](/solution/2600-2699/2629.Function%20Composition/README.md)                                                                                 |      | 简单 |      |
 | 2630 | [记忆函数 II](/solution/2600-2699/2630.Memoize%20II/README.md)                                                                                        |      | 困难 |      |
 | 2631 | [分组](/solution/2600-2699/2631.Group%20By/README.md)                                                                                                 |      | 中等 |      |
-| 2632 | [柯里化](/solution/2600-2699/2632.Curry/README.md)                                                                                                    |      | 困难 | 🔒   |
+| 2632 | [柯里化](/solution/2600-2699/2632.Curry/README.md)                                                                                                    |      | 中等 | 🔒   |
 | 2633 | [将对象转换为 JSON 字符串](/solution/2600-2699/2633.Convert%20Object%20to%20JSON%20String/README.md)                                                  |      | 中等 | 🔒   |
 | 2634 | [过滤数组中的元素](/solution/2600-2699/2634.Filter%20Elements%20from%20Array/README.md)                                                               |      | 简单 |      |
 | 2635 | [转换数组中的每个元素](/solution/2600-2699/2635.Apply%20Transform%20Over%20Each%20Element%20in%20Array/README.md)                                     |      | 简单 |      |

solution/JAVASCRIPT_README_EN.md

@@ -22,7 +22,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 | 2629 | [Function Composition](/solution/2600-2699/2629.Function%20Composition/README_EN.md)                                                                                                   |      | Easy       |        |
 | 2630 | [Memoize II](/solution/2600-2699/2630.Memoize%20II/README_EN.md)                                                                                                                       |      | Hard       |        |
 | 2631 | [Group By](/solution/2600-2699/2631.Group%20By/README_EN.md)                                                                                                                           |      | Medium     |        |
-| 2632 | [Curry](/solution/2600-2699/2632.Curry/README_EN.md)                                                                                                                                   |      | Hard       | 🔒     |
+| 2632 | [Curry](/solution/2600-2699/2632.Curry/README_EN.md)                                                                                                                                   |      | Medium     | 🔒     |
 | 2633 | [Convert Object to JSON String](/solution/2600-2699/2633.Convert%20Object%20to%20JSON%20String/README_EN.md)                                                                           |      | Medium     | 🔒     |
 | 2634 | [Filter Elements from Array](/solution/2600-2699/2634.Filter%20Elements%20from%20Array/README_EN.md)                                                                                   |      | Easy       |        |
 | 2635 | [Apply Transform Over Each Element in Array](/solution/2600-2699/2635.Apply%20Transform%20Over%20Each%20Element%20in%20Array/README_EN.md)                                             |      | Easy       |        |