feat: add solutions to lc problem: No.0829

solution/0800-0899/0829.Consecutive Numbers Sum/README.md

@@ -58,15 +58,21 @@ tags:
 
 ### 方法一：数学推导
 
-连续正整数构成一个等差数列($d=1$)。我们假设等差数列的第一项为 $a$，项数为 $k$，则 $n=(a+a+k-1)*k/2$，即 $n*2=(a*2+k-1)*k$ ①。
+连续正整数构成一个公差 $d = 1$ 的等差数列。我们假设等差数列的第一项为 $a$，项数为 $k$，那么 $n = (a + a + k - 1) \times k / 2$，即 $n \times 2 = (a \times 2 + k - 1) \times k$。这里我们可以得出 $k$ 一定能整除 $n \times 2$，并且 $(n \times 2) / k - k + 1$ 一定是偶数。
 
-由于是连续正整数， $a>=1$，所以 ① 可以化为 $n*2>=(k+1)*k$，即 $k*(k+1)<=n*2$ ②。
+由于 $a \geq 1$，所以 $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$。
 
-因此，$k$ 的范围需要满足 $k>=1$ 并且 $k*(k+1)<=n*2$。另外，我们从 ① 式可以发现，$k$ 必须能整除 $n*2$。
+综上，我们可以得出：
 
-综上，我们枚举 $k$，累加满足条件的 $k$ 的个数即可。
+1. $k$ 一定能整除 $n \times 2$；
+2. $k \times (k + 1) \leq n \times 2$；
+3. $(n \times 2) / k - k + 1$ 一定是偶数。
 
-时间复杂度 $O(\sqrt{n})$。
+我们从 $k = 1$ 开始枚举，当 $k \times (k + 1) > n \times 2$ 时，我们可以结束枚举。在枚举的过程中，我们判断 $k$ 是否能整除 $n \times 2$，并且 $(n \times 2) / k - k + 1$ 是否是偶数，如果是则满足条件，答案加一。
+
+枚举结束后，返回答案即可。
+
+时间复杂度 $O(\sqrt{n})$，其中 $n$ 为给定的正整数。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -135,6 +141,21 @@ func consecutiveNumbersSum(n int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function consecutiveNumbersSum(n: number): number {
+    let ans = 0;
+    n <<= 1;
+    for (let k = 1; k * (k + 1) <= n; ++k) {
+        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+            ++ans;
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0800-0899/0829.Consecutive Numbers Sum/README_EN.md

@@ -57,7 +57,23 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Mathematical Derivation
+
+Consecutive positive integers form an arithmetic sequence with a common difference $d = 1$. Let's assume the first term of the sequence is $a$, and the number of terms is $k$. Then, $n = (a + a + k - 1) \times k / 2$, which simplifies to $n \times 2 = (a \times 2 + k - 1) \times k$. From this, we can deduce that $k$ must divide $n \times 2$ evenly, and $(n \times 2) / k - k + 1$ must be an even number.
+
+Given that $a \geq 1$, it follows that $n \times 2 = (a \times 2 + k - 1) \times k \geq k \times (k + 1)$.
+
+In summary, we can conclude:
+
+1. $k$ must divide $n \times 2$ evenly;
+2. $k \times (k + 1) \leq n \times 2$;
+3. $(n \times 2) / k - k + 1$ must be an even number.
+
+We start enumerating from $k = 1$, and we can stop when $k \times (k + 1) > n \times 2$. During the enumeration, we check if $k$ divides $n \times 2$ evenly, and if $(n \times 2) / k - k + 1$ is an even number. If both conditions are met, it satisfies the criteria, and we increment the answer by one.
+
+After finishing the enumeration, we return the answer.
+
+The time complexity is $O(\sqrt{n})$, where $n$ is the given positive integer. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -69,7 +85,7 @@ class Solution:
         n <<= 1
         ans, k = 0, 1
         while k * (k + 1) <= n:
-            if n % k == 0 and (n // k + 1 - k) % 2 == 0:
+            if n % k == 0 and (n // k - k + 1) % 2 == 0:
                 ans += 1
             k += 1
         return ans
@@ -126,6 +142,21 @@ func consecutiveNumbersSum(n int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function consecutiveNumbersSum(n: number): number {
+    let ans = 0;
+    n <<= 1;
+    for (let k = 1; k * (k + 1) <= n; ++k) {
+        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+            ++ans;
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0800-0899/0829.Consecutive Numbers Sum/Solution.py

@@ -3,7 +3,7 @@ def consecutiveNumbersSum(self, n: int) -> int:
         n <<= 1
         ans, k = 0, 1
         while k * (k + 1) <= n:
-            if n % k == 0 and (n // k + 1 - k) % 2 == 0:
+            if n % k == 0 and (n // k - k + 1) % 2 == 0:
                 ans += 1
             k += 1
         return ans

solution/0800-0899/0829.Consecutive Numbers Sum/Solution.ts

@@ -0,0 +1,10 @@
+function consecutiveNumbersSum(n: number): number {
+    let ans = 0;
+    n <<= 1;
+    for (let k = 1; k * (k + 1) <= n; ++k) {
+        if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
+            ++ans;
+        }
+    }
+    return ans;
+}

solution/3100-3199/3188.Find Top Scoring Students II/README.md

@@ -6,15 +6,15 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Fi
 
 <!-- problem:start -->
 
-# [3188. Find Top Scoring Students II 🔒](https://leetcode.cn/problems/find-top-scoring-students-ii)
+# [3188. 查找得分最高的学生 II 🔒](https://leetcode.cn/problems/find-top-scoring-students-ii)
 
 [English Version](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README_EN.md)
 
 ## 题目描述
 
 <!-- description:start -->
 
-<p>Table: <code>students</code></p>
+<p>表：<code>students</code></p>
 
 <pre>
 +-------------+----------+
@@ -24,11 +24,11 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Fi
 | name        | varchar  |
 | major       | varchar  |
 +-------------+----------+
-student_id is the primary key for this table. 
-Each row contains the student ID, student name, and their major.
+student_id 是这张表的主键（有不同值的列的组合）。
+这张表的每一行包含学生 ID，学生姓名和他们的专业。
 </pre>
 
-<p>Table: <code>courses</code></p>
+<p>表：<code>courses</code></p>
 
 <pre>
 +-------------+-------------------+
@@ -40,12 +40,12 @@ Each row contains the student ID, student name, and their major.
 | major       | varchar           |       
 | mandatory   | enum              |      
 +-------------+-------------------+
-course_id is the primary key for this table. 
-mandatory is an enum type of (&#39;Yes&#39;, &#39;No&#39;).
-Each row contains the course ID, course name, credits, major it belongs to, and whether the course is mandatory.
+course_id 是这张表的主键。 
+mandatory 是 ('Yes', 'No') 的枚举类型。
+每一行包含课程 ID，课程名，学分，所属专业，以及该课程是否必修。
 </pre>
 
-<p>Table: <code>enrollments</code></p>
+<p>表：<code>enrollments</code></p>
 
 <pre>
 +-------------+----------+
@@ -57,27 +57,28 @@ Each row contains the course ID, course name, credits, major it belongs to, and
 | grade       | varchar  |
 | GPA         | decimal  | 
 +-------------+----------+
-(student_id, course_id, semester) is the primary key (combination of columns with unique values) for this table.
-Each row contains the student ID, course ID, semester, and grade received.
+(student_id, course_id, semester) 是这张表的主键（有不同值的列的组合）。
+这张表的每一行包含学生 ID，课程 ID，学期和获得的学分。
 </pre>
 
-<p>Write a solution to find the students who meet the following criteria:</p>
+<p>编写一个解决方案来查找满足下述标准的学生：</p>
 
 <ul>
-	<li>Have<strong> taken all mandatory courses</strong> and <strong>at least two</strong> elective courses offered in <strong>their major.</strong></li>
-	<li>Achieved a grade of <strong>A</strong>&nbsp;in <strong>all mandatory courses</strong> and at least <strong>B</strong>&nbsp;in<strong> elective courses</strong>.</li>
-	<li>Maintained an average <code>GPA</code> of at least&nbsp;<code>2.5</code> across all their courses (including those outside their major).</li>
+	<li>已经 <strong>修完他们专业中所有的必修课程</strong> 和 <strong>至少两个&nbsp;</strong>选修课程。</li>
+	<li>在 <strong>所有必修课程</strong> 中取得等级 <strong>A</strong> 并且 <strong>选修课程</strong> 至少取得 <strong>B</strong>。</li>
+	<li>保持他们所有课程（包括不属于他们专业的）的平均&nbsp;<code>GPA</code>&nbsp;至少在&nbsp;<code>2.5</code>&nbsp;以上。</li>
 </ul>
 
-<p>Return <em>the result table ordered by</em> <code>student_id</code> <em>in <strong>ascending</strong> order</em>.</p>
+<p>返回结果表以&nbsp;<code>student_id</code> <strong>升序&nbsp;</strong>排序。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example:</strong></p>
+
+<p><strong class="example">示例：</strong></p>
 
 <div class="example-block">
-<p><strong>Input:</strong></p>
+<p><strong>输入：</strong></p>
 
-<p>students table:</p>
+<p>students 表：</p>
 
 <pre class="example-io">
  +------------+------------------+------------------+
@@ -90,7 +91,7 @@ Each row contains the student ID, course ID, semester, and grade received.
  +------------+------------------+------------------+
  </pre>
 
-<p>courses table:</p>
+<p>courses 表：</p>
 
 <pre class="example-io">
  +-----------+-------------------+---------+------------------+----------+
@@ -107,7 +108,7 @@ Each row contains the student ID, course ID, semester, and grade received.
  +-----------+-------------------+---------+------------------+----------+
  </pre>
 
-<p>enrollments table:</p>
+<p>enrollments 表：</p>
 
 <pre class="example-io">
  +------------+-----------+-------------+-------+-----+
@@ -128,7 +129,7 @@ Each row contains the student ID, course ID, semester, and grade received.
  +------------+-----------+-------------+-------+-----+
  </pre>
 
-<p><strong>Output:</strong></p>
+<p><strong>输出：</strong></p>
 
 <pre class="example-io">
  +------------+
@@ -139,16 +140,16 @@ Each row contains the student ID, course ID, semester, and grade received.
  +------------+
  </pre>
 
-<p><strong>Explanation:</strong></p>
+<p><strong>解释：</strong></p>
 
 <ul>
-	<li>Alice (student_id 1) is a Computer Science major and has taken both Algorithms&nbsp;and Data Structures, receiving an A&nbsp;in both. She has also taken Machine Learning&nbsp;and Operating Systems&nbsp;as electives, receiving an A&nbsp;and B&nbsp;respectively.</li>
-	<li>Bob (student_id 2) is a Computer Science major but did not receive an A&nbsp;in all required courses.</li>
-	<li>Charlie (student_id 3) is a Mathematics major and has taken both Calculus&nbsp;and Linear Algebra, receiving an A&nbsp;in both. He has also taken Probability&nbsp;and Statistics&nbsp;as electives, receiving an A&nbsp;and B&nbsp;respectively.</li>
-	<li>David (student_id 4) is a Mathematics major but did not receive an A&nbsp;in all required courses.</li>
+	<li>Alice (student_id 1) 是计算机科学专业并且修了&nbsp;Algorithms&nbsp;和 Data Structures，都取得了 A。她同时选修了&nbsp;Machine Learning&nbsp;和 Operating Systems，分别取得了 A 和 B。</li>
+	<li>Bob (student_id 2) 是计算机科学专业但没有在所有需求的课程中取得 A。</li>
+	<li>Charlie (student_id 3) 是数学专业并且修了 Calculus&nbsp;和 Linear Algebra，都取得了 A。他同时选修了&nbsp;Probability&nbsp;和 Statistics，分别取得了 A 和 B。</li>
+	<li>David (student_id 4) 是数学专业但没有在所有需要的课程中取得 A。</li>
 </ul>
 
-<p><strong>Note:</strong> Output table is ordered by student_id in ascending order.</p>
+<p><strong>注意：</strong>输出表以 student_id 升序排序。</p>
 </div>
 
 <!-- description:end -->
@@ -159,11 +160,11 @@ Each row contains the student ID, course ID, semester, and grade received.
 
 ### 方法一：连接 + 分组 + 条件过滤
 
-我们首先计算出每个学生的平均 GPA，记录在临时表 `T` 中。
+我们首先筛选出平均 GPA 大于等于 2.5 的学生，记录在 `T` 表中。
 
-然后，我们将 `students` 表与 `courses` 表按照 `major` 进行连接，然后与 `T` 表按照 `student_id` 进行连接，再与 `enrollments` 表按照 `student_id` 和 `course_id` 进行左连接。
+然后，我们将 `T` 表与 `students` 表按照 `student_id` 进行连接，然后与 `courses` 表按照 `major` 进行连接，再与 `enrollments` 表按照 `student_id` 和 `course_id` 进行左连接。
 
-接下来，我们筛选出平均 GPA 大于等于 2.5 的学生，并按照学生 ID 进行分组，然后使用 `HAVING` 子句过滤出符合条件的学生，最后按照学生 ID 进行排序。
+接下来，我们按照学生 ID 进行分组，然后使用 `HAVING` 子句过滤出符合条件的学生，最后按照学生 ID 进行排序。
 
 <!-- tabs:start -->
 
@@ -173,17 +174,17 @@ Each row contains the student ID, course ID, semester, and grade received.
 # Write your MySQL query statement below
 WITH
     T AS (
-        SELECT student_id, AVG(GPA) AS avg_gpa
+        SELECT student_id
         FROM enrollments
         GROUP BY 1
+        HAVING AVG(GPA) >= 2.5
     )
 SELECT student_id
 FROM
-    students
+    T
+    JOIN students USING (student_id)
     JOIN courses USING (major)
-    JOIN T USING (student_id)
     LEFT JOIN enrollments USING (student_id, course_id)
-WHERE avg_gpa >= 2.5
 GROUP BY 1
 HAVING
     SUM(mandatory = 'yes' AND grade = 'A') = SUM(mandatory = 'yes')

solution/3100-3199/3188.Find Top Scoring Students II/README_EN.md

@@ -159,11 +159,11 @@ Each row contains the student ID, course ID, semester, and grade received.
 
 ### Solution 1: Joining + Grouping + Conditional Filtering
 
-First, we calculate the average GPA of each student and store it in a temporary table `T`.
+First, we filter out students with an average GPA greater than or equal to 2.5 and record them in table `T`.
 
-Next, we join the `students` table with the `courses` table based on `major`, and then join with the `T` table based on `student_id`, followed by a left join with the `enrollments` table based on `student_id` and `course_id`.
+Next, we join the `T` table with the `students` table based on `student_id`, then join with the `courses` table based on `major`, and finally perform a left join with the `enrollments` table based on `student_id` and `course_id`.
 
-After that, we filter out students with an average GPA greater than or equal to 2.5, group by student ID, use the `HAVING` clause to filter students who meet the criteria, and finally sort by student ID.
+After that, we group by student ID, use the `HAVING` clause to filter out students who meet the conditions, and finally sort by student ID.
 
 <!-- tabs:start -->
 
@@ -173,17 +173,17 @@ After that, we filter out students with an average GPA greater than or equal to
 # Write your MySQL query statement below
 WITH
     T AS (
-        SELECT student_id, AVG(GPA) AS avg_gpa
+        SELECT student_id
         FROM enrollments
         GROUP BY 1
+        HAVING AVG(GPA) >= 2.5
     )
 SELECT student_id
 FROM
-    students
+    T
+    JOIN students USING (student_id)
     JOIN courses USING (major)
-    JOIN T USING (student_id)
     LEFT JOIN enrollments USING (student_id, course_id)
-WHERE avg_gpa >= 2.5
 GROUP BY 1
 HAVING
     SUM(mandatory = 'yes' AND grade = 'A') = SUM(mandatory = 'yes')

solution/3100-3199/3188.Find Top Scoring Students II/Solution.sql

@@ -1,17 +1,17 @@
 # Write your MySQL query statement below
 WITH
     T AS (
-        SELECT student_id, AVG(GPA) AS avg_gpa
+        SELECT student_id
         FROM enrollments
         GROUP BY 1
+        HAVING AVG(GPA) >= 2.5
     )
 SELECT student_id
 FROM
-    students
+    T
+    JOIN students USING (student_id)
     JOIN courses USING (major)
-    JOIN T USING (student_id)
     LEFT JOIN enrollments USING (student_id, course_id)
-WHERE avg_gpa >= 2.5
 GROUP BY 1
 HAVING
     SUM(mandatory = 'yes' AND grade = 'A') = SUM(mandatory = 'yes')

solution/DATABASE_README.md

@@ -283,7 +283,7 @@
 | 3166 | [计算停车费与时长](/solution/3100-3199/3166.Calculate%20Parking%20Fees%20and%20Duration/README.md)                                                           | `数据库` | 中等 | 🔒   |
 | 3172 | [第二天验证](/solution/3100-3199/3172.Second%20Day%20Verification/README.md)                                                                                 | `数据库` | 简单 | 🔒   |
 | 3182 | [查找得分最高的学生](/solution/3100-3199/3182.Find%20Top%20Scoring%20Students/README.md)                                                                     | `数据库` | 中等 | 🔒   |
-| 3188 | [Find Top Scoring Students II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md)                                                      |          | 困难 | 🔒   |
+| 3188 | [查找得分最高的学生 II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md)                                                             |          | 困难 | 🔒   |
 
 ## 版权
 

solution/README.md

@@ -3198,7 +3198,7 @@
 |  3185  |  [构成整天的下标对数目 II](/solution/3100-3199/3185.Count%20Pairs%20That%20Form%20a%20Complete%20Day%20II/README.md)  |  `数组`,`哈希表`,`计数`  |  中等  |  第 402 场周赛  |
 |  3186  |  [施咒的最大总伤害](/solution/3100-3199/3186.Maximum%20Total%20Damage%20With%20Spell%20Casting/README.md)  |  `数组`,`哈希表`,`双指针`,`二分查找`,`动态规划`,`计数`,`排序`  |  中等  |  第 402 场周赛  |
 |  3187  |  [数组中的峰值](/solution/3100-3199/3187.Peaks%20in%20Array/README.md)  |  `树状数组`,`线段树`,`数组`  |  困难  |  第 402 场周赛  |
-|  3188  |  [Find Top Scoring Students II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md)  |    |  困难  |  🔒  |
+|  3188  |  [查找得分最高的学生 II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md)  |    |  困难  |  🔒  |
 
 ## 版权