feat: add solutions to lc problem: No.3278

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/README.md

@@ -314,7 +314,7 @@ function getDirections(root: TreeNode | null, startValue: number, destValue: num
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
@@ -370,7 +370,7 @@ var getDirections = function (root, startValue, destValue) {
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node, x, path) => {

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/README_EN.md

@@ -310,7 +310,7 @@ function getDirections(root: TreeNode | null, startValue: number, destValue: num
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
@@ -366,7 +366,7 @@ var getDirections = function (root, startValue, destValue) {
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node, x, path) => {

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/Solution.js

@@ -20,7 +20,7 @@ var getDirections = function (root, startValue, destValue) {
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node, x, path) => {

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/Solution.ts

@@ -20,7 +20,7 @@ function getDirections(root: TreeNode | null, startValue: number, destValue: num
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {

solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/README.md

@@ -66,6 +66,10 @@ startDate = 2022-03-08, endDate = 2022-03-20, minAmount = 1000
  - 用户 2 在时间间隔内有一次购买，但金额小于 minAmount。
  - 用户 3 是唯一一个购买行为同时满足这两个条件的用户。</pre>
 
+<p>&nbsp;</p>
+
+<p><b>重要提示：</b>此问题与 <a href="https://leetcode.cn/problems/the-users-that-are-eligible-for-discount/">2230. 查找可享受优惠的用户</a>&nbsp;基本相同。</p>
+
 <!-- description:end -->
 
 ## 解法

solution/3200-3299/3278.Find Candidates for Data Scientist Position II/README.md

@@ -0,0 +1,197 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3278.Find%20Candidates%20for%20Data%20Scientist%20Position%20II/README.md
+tags:
+    - 数据库
+---
+
+<!-- problem:start -->
+
+# [3278. Find Candidates for Data Scientist Position II 🔒](https://leetcode.cn/problems/find-candidates-for-data-scientist-position-ii)
+
+[English Version](/solution/3200-3299/3278.Find%20Candidates%20for%20Data%20Scientist%20Position%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>Table: <font face="monospace"><code>Candidates</code></font></p>
+
+<pre>
++--------------+---------+ 
+| Column Name  | Type    | 
++--------------+---------+ 
+| candidate_id | int     | 
+| skill        | varchar |
+| proficiency  | int     |
++--------------+---------+
+(candidate_id, skill) is the unique key for this table.
+Each row includes candidate_id, skill, and proficiency level (1-5).
+</pre>
+
+<p>Table: <font face="monospace"><code>Projects</code></font></p>
+
+<pre>
++--------------+---------+ 
+| Column Name  | Type    | 
++--------------+---------+ 
+| project_id   | int     | 
+| skill        | varchar |
+| importance   | int     |
++--------------+---------+
+(project_id, skill) is the primary key for this table.
+Each row includes project_id, required skill, and its importance (1-5) for the project.
+</pre>
+
+<p>Leetcode is staffing for multiple data science projects. Write a solution to find the <strong>best candidate</strong> for<strong> each project</strong> based on the following criteria:</p>
+
+<ol>
+	<li>Candidates must have <strong>all</strong> the skills required for a project.</li>
+	<li>Calculate a <strong>score</strong> for each candidate-project pair as follows:
+	<ul>
+		<li><strong>Start</strong> with <code>100</code> points</li>
+		<li><strong>Add</strong> <code>10</code> points for each skill where <strong>proficiency &gt; importance</strong></li>
+		<li><strong>Subtract</strong> <code>5</code> points for each skill where <strong>proficiency &lt; importance</strong></li>
+	</ul>
+	</li>
+</ol>
+
+<p>Include only the top candidate (highest score) for each project. If there&rsquo;s a <strong>tie</strong>, choose the candidate with the <strong>lower</strong> <code>candidate_id</code>. If there is <strong>no suitable candidate</strong> for a project, <strong>do not return</strong>&nbsp;that project.</p>
+
+<p>Return a result table ordered by <code>project_id</code> in ascending order.</p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong></p>
+
+<p><code>Candidates</code> table:</p>
+
+<pre class="example-io">
++--------------+-----------+-------------+
+| candidate_id | skill     | proficiency |
++--------------+-----------+-------------+
+| 101          | Python    | 5           |
+| 101          | Tableau   | 3           |
+| 101          | PostgreSQL| 4           |
+| 101          | TensorFlow| 2           |
+| 102          | Python    | 4           |
+| 102          | Tableau   | 5           |
+| 102          | PostgreSQL| 4           |
+| 102          | R         | 4           |
+| 103          | Python    | 3           |
+| 103          | Tableau   | 5           |
+| 103          | PostgreSQL| 5           |
+| 103          | Spark     | 4           |
++--------------+-----------+-------------+
+</pre>
+
+<p><code>Projects</code> table:</p>
+
+<pre class="example-io">
++-------------+-----------+------------+
+| project_id  | skill     | importance |
++-------------+-----------+------------+
+| 501         | Python    | 4          |
+| 501         | Tableau   | 3          |
+| 501         | PostgreSQL| 5          |
+| 502         | Python    | 3          |
+| 502         | Tableau   | 4          |
+| 502         | R         | 2          |
++-------------+-----------+------------+
+</pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++-------------+--------------+-------+
+| project_id  | candidate_id | score |
++-------------+--------------+-------+
+| 501         | 101          | 105   |
+| 502         | 102          | 130   |
++-------------+--------------+-------+
+</pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For Project 501, Candidate 101 has the highest score of 105. All other candidates have the same score but Candidate 101 has the lowest candidate_id among them.</li>
+	<li>For Project 502, Candidate 102 has the highest score of 130.</li>
+</ul>
+
+<p>The output table is ordered by project_id in ascending order.</p>
+</div>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：等值连接 + 分组统计 + 窗口函数
+
+我们可以将表 `Candidates` 和表 `Projects` 通过 `skill` 列进行等值连接，统计每个候选人在每个项目中匹配的技能数量、总分数，记录在表 `S` 中。
+
+然后我们再次统计每个项目所需的技能数量，记录在表 `T` 中。
+
+接着我们将表 `S` 和表 `T` 通过 `project_id` 列进行等值连接，筛选出匹配的技能数量等于所需技能数量的候选人，记录在表 `P` 中，并计算每个项目的候选人排名，字段为 `rk`。
+
+最后我们筛选出每个项目的排名为 1 的候选人，即为最佳候选人。
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+WITH
+    S AS (
+        SELECT
+            candidate_id,
+            project_id,
+            COUNT(*) matched_skills,
+            SUM(
+                CASE
+                    WHEN proficiency > importance THEN 10
+                    WHEN proficiency < importance THEN -5
+                    ELSE 0
+                END
+            ) + 100 AS score
+        FROM
+            Candidates
+            JOIN Projects USING (skill)
+        GROUP BY 1, 2
+    ),
+    T AS (
+        SELECT project_id, COUNT(1) required_skills
+        FROM Projects
+        GROUP BY 1
+    ),
+    P AS (
+        SELECT
+            project_id,
+            candidate_id,
+            score,
+            RANK() OVER (
+                PARTITION BY project_id
+                ORDER BY score DESC, candidate_id
+            ) rk
+        FROM
+            S
+            JOIN T USING (project_id)
+        WHERE matched_skills = required_skills
+    )
+SELECT project_id, candidate_id, score
+FROM P
+WHERE rk = 1
+ORDER BY 1;
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->