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feat: add solutions to lc problem: No.3278 (doocs#3467)
No.3278.Find Candidates for Data Scientist Position II
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‎solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/README.md

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

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

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const left = lca(node.left, p, q);
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const right = lca(node.right, p, q);
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return left && right ? node : left ?? right;
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return left && right ? node : (left ?? right);
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};
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const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
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const left = lca(node.left, p, q);
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const right = lca(node.right, p, q);
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return left && right ? node : left ?? right;
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return left && right ? node : (left ?? right);
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};
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const dfs = (node, x, path) => {

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

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const left = lca(node.left, p, q);
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const right = lca(node.right, p, q);
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return left && right ? node : left ?? right;
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return left && right ? node : (left ?? right);
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};
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const dfs = (node, x, path) => {

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

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const left = lca(node.left, p, q);
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const right = lca(node.right, p, q);
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return left && right ? node : left ?? right;
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return left && right ? node : (left ?? right);
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};
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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

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- 用户 2 在时间间隔内有一次购买，但金额小于 minAmount。
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- 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>
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<p>&nbsp;</p>
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<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>
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<!-- description:end -->
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## 解法

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

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3278.Find%20Candidates%20for%20Data%20Scientist%20Position%20II/README.md
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tags:
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- 数据库
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---
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<!-- problem:start -->
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# [3278. Find Candidates for Data Scientist Position II 🔒](https://leetcode.cn/problems/find-candidates-for-data-scientist-position-ii)
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[English Version](/solution/3200-3299/3278.Find%20Candidates%20for%20Data%20Scientist%20Position%20II/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Table: <font face="monospace"><code>Candidates</code></font></p>
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<pre>
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+--------------+---------+
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| Column Name | Type |
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+--------------+---------+
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| candidate_id | int |
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| skill | varchar |
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| proficiency | int |
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+--------------+---------+
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(candidate_id, skill) is the unique key for this table.
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Each row includes candidate_id, skill, and proficiency level (1-5).
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</pre>
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<p>Table: <font face="monospace"><code>Projects</code></font></p>
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<pre>
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+--------------+---------+
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| Column Name | Type |
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+--------------+---------+
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| project_id | int |
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| skill | varchar |
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| importance | int |
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+--------------+---------+
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(project_id, skill) is the primary key for this table.
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Each row includes project_id, required skill, and its importance (1-5) for the project.
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</pre>
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<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>
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<ol>
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<li>Candidates must have <strong>all</strong> the skills required for a project.</li>
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<li>Calculate a <strong>score</strong> for each candidate-project pair as follows:
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<ul>
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<li><strong>Start</strong> with <code>100</code> points</li>
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<li><strong>Add</strong> <code>10</code> points for each skill where <strong>proficiency &gt; importance</strong></li>
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<li><strong>Subtract</strong> <code>5</code> points for each skill where <strong>proficiency &lt; importance</strong></li>
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</ul>
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</li>
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</ol>
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<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>
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<p>Return a result table ordered by <code>project_id</code> in ascending order.</p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><code>Candidates</code> table:</p>
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<pre class="example-io">
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+--------------+-----------+-------------+
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| candidate_id | skill | proficiency |
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+--------------+-----------+-------------+
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| 101 | Python | 5 |
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| 101 | Tableau | 3 |
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| 101 | PostgreSQL| 4 |
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| 101 | TensorFlow| 2 |
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| 102 | Python | 4 |
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| 102 | Tableau | 5 |
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| 102 | PostgreSQL| 4 |
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| 102 | R | 4 |
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| 103 | Python | 3 |
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| 103 | Tableau | 5 |
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| 103 | PostgreSQL| 5 |
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| 103 | Spark | 4 |
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+--------------+-----------+-------------+
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</pre>
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<p><code>Projects</code> table:</p>
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<pre class="example-io">
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+-------------+-----------+------------+
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| project_id | skill | importance |
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+-------------+-----------+------------+
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| 501 | Python | 4 |
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| 501 | Tableau | 3 |
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| 501 | PostgreSQL| 5 |
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| 502 | Python | 3 |
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| 502 | Tableau | 4 |
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| 502 | R | 2 |
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+-------------+-----------+------------+
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</pre>
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<p><strong>Output:</strong></p>
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<pre class="example-io">
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+-------------+--------------+-------+
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| project_id | candidate_id | score |
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+-------------+--------------+-------+
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| 501 | 101 | 105 |
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| 502 | 102 | 130 |
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+-------------+--------------+-------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<ul>
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<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>
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<li>For Project 502, Candidate 102 has the highest score of 130.</li>
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</ul>
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<p>The output table is ordered by project_id in ascending order.</p>
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</div>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一：等值连接 + 分组统计 + 窗口函数
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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。
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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。
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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`。
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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。
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<!-- tabs:start -->
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#### MySQL
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```sql
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WITH
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S AS (
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SELECT
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candidate_id,
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project_id,
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COUNT(*) matched_skills,
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SUM(
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CASE
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WHEN proficiency > importance THEN 10
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WHEN proficiency < importance THEN -5
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ELSE 0
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END
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) + 100 AS score
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FROM
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Candidates
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JOIN Projects USING (skill)
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GROUP BY 1, 2
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),
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T AS (
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SELECT project_id, COUNT(1) required_skills
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FROM Projects
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GROUP BY 1
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),
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P AS (
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SELECT
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project_id,
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candidate_id,
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score,
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RANK() OVER (
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PARTITION BY project_id
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ORDER BY score DESC, candidate_id
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) rk
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FROM
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S
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JOIN T USING (project_id)
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WHERE matched_skills = required_skills
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)
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SELECT project_id, candidate_id, score
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FROM P
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WHERE rk = 1
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ORDER BY 1;
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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