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Commit 975b81e

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Added task 3580
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3580\. Find Consistently Improving Employees
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Medium
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Table: `employees`
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| employee_id | int |
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| name | varchar |
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+-------------+---------+
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employee_id is the unique identifier for this table.
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Each row contains information about an employee.
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Table: `performance_reviews`
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+-------------+------+
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| Column Name | Type |
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+-------------+------+
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| review_id | int |
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| employee_id | int |
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| review_date | date |
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| rating | int |
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+-------------+------+
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review_id is the unique identifier for this table.
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Each row represents a performance review for an employee.
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The rating is on a scale of 1-5 where 5 is excellent and 1 is poor.
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Write a solution to find employees who have consistently improved their performance over **their last three reviews**.
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* An employee must have **at least** `3` **review** to be considered
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* The employee's **last** `3` **reviews** must show **strictly increasing ratings** (each review better than the previous)
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* Use the most recent `3` reviews based on `review_date` for each employee
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* Calculate the **improvement score** as the difference between the latest rating and the earliest rating among the last `3` reviews
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Return _the result table ordered by **improvement score** in **descending** order, then by **name** in **ascending** order_.
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The result format is in the following example.
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**Example:**
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**Input:**
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employees table:
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+-------------+----------------+
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| employee_id | name |
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+-------------+----------------+
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| 1 | Alice Johnson |
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| 2 | Bob Smith |
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| 3 | Carol Davis |
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| 4 | David Wilson |
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| 5 | Emma Brown |
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+-------------+----------------+
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performance\_reviews table:
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+-----------+-------------+-------------+--------+
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| review_id | employee_id | review_date | rating |
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+-----------+-------------+-------------+--------+
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| 1 | 1 | 2023年01月15日 | 2 |
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| 2 | 1 | 2023年04月15日 | 3 |
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| 3 | 1 | 2023年07月15日 | 4 |
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| 4 | 1 | 2023年10月15日 | 5 |
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| 5 | 2 | 2023年02月01日 | 3 |
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| 6 | 2 | 2023年05月01日 | 2 |
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| 7 | 2 | 2023年08月01日 | 4 |
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| 8 | 2 | 2023年11月01日 | 5 |
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| 9 | 3 | 2023年03月10日 | 1 |
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| 10 | 3 | 2023年06月10日 | 2 |
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| 11 | 3 | 2023年09月10日 | 3 |
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| 12 | 3 | 2023年12月10日 | 4 |
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| 13 | 4 | 2023年01月20日 | 4 |
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| 14 | 4 | 2023年04月20日 | 4 |
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| 15 | 4 | 2023年07月20日 | 4 |
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| 16 | 5 | 2023年02月15日 | 3 |
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| 17 | 5 | 2023年05月15日 | 2 |
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+-----------+-------------+-------------+--------+
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**Output:**
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+-------------+----------------+-------------------+
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| employee_id | name | improvement_score |
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+-------------+----------------+-------------------+
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| 2 | Bob Smith | 3 |
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| 1 | Alice Johnson | 2 |
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| 3 | Carol Davis | 2 |
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+-------------+----------------+-------------------+
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**Explanation:**
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* **Alice Johnson (employee\_id = 1):**
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* Has 4 reviews with ratings: 2, 3, 4, 5
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* Last 3 reviews (by date): 2023年04月15日 (3), 2023年07月15日 (4), 2023年10月15日 (5)
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* Ratings are strictly increasing: 3 → 4 → 5
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* Improvement score: 5 - 3 = 2
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* **Carol Davis (employee\_id = 3):**
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* Has 4 reviews with ratings: 1, 2, 3, 4
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* Last 3 reviews (by date): 2023年06月10日 (2), 2023年09月10日 (3), 2023年12月10日 (4)
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* Ratings are strictly increasing: 2 → 3 → 4
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* Improvement score: 4 - 2 = 2
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* **Bob Smith (employee\_id = 2):**
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* Has 4 reviews with ratings: 3, 2, 4, 5
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* Last 3 reviews (by date): 2023年05月01日 (2), 2023年08月01日 (4), 2023年11月01日 (5)
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* Ratings are strictly increasing: 2 → 4 → 5
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* Improvement score: 5 - 2 = 3
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* **Employees not included:**
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* David Wilson (employee\_id = 4): Last 3 reviews are all 4 (no improvement)
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* Emma Brown (employee\_id = 5): Only has 2 reviews (needs at least 3)
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The output table is ordered by improvement\_score in descending order, then by name in ascending order.
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# Write your MySQL query statement below
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# #Medium #Database #2025_06_11_Time_449_ms_(91.67%)_Space_0.0_MB_(100.00%)
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WITH Ranked AS (
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SELECT
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e.employee_id,
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e.name,
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pr.review_date,
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pr.rating,
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RANK() OVER (
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PARTITION BY e.employee_id
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ORDER BY pr.review_date DESC
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) AS rnk,
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LAG(pr.rating) OVER (
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PARTITION BY e.employee_id
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ORDER BY pr.review_date DESC
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) AS lag_rating
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FROM employees e
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LEFT JOIN performance_reviews pr
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ON e.employee_id = pr.employee_id
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)
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SELECT
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employee_id,
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name,
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MAX(rating) - MIN(rating) AS improvement_score
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FROM Ranked
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WHERE rnk <= 3
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GROUP BY
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employee_id,
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name
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HAVING
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COUNT(*) = 3
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AND SUM(CASE WHEN lag_rating > rating THEN 1 ELSE 0 END) = 2
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ORDER BY
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improvement_score DESC,
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name ASC;
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package g3501_3600.s3580_find_consistently_improving_employees;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import java.io.BufferedReader;
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import java.io.FileNotFoundException;
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import java.io.FileReader;
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import java.sql.Connection;
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import java.sql.ResultSet;
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import java.sql.SQLException;
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import java.sql.Statement;
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import java.util.stream.Collectors;
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import javax.sql.DataSource;
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import org.junit.jupiter.api.Test;
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import org.zapodot.junit.db.annotations.EmbeddedDatabase;
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import org.zapodot.junit.db.annotations.EmbeddedDatabaseTest;
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import org.zapodot.junit.db.common.CompatibilityMode;
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@EmbeddedDatabaseTest(
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compatibilityMode = CompatibilityMode.MySQL,
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initialSqls =
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"CREATE TABLE employees(employee_id INTEGER, name VARCHAR(255)); "
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+ "INSERT INTO employees (employee_id, name) VALUES"
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+ " (1, 'Alice Johnson'),"
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+ " (2, 'Bob Smith'),"
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+ " (3, 'Carol Davis'),"
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+ " (4, 'David Wilson'),"
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+ " (5, 'Emma Brown');"
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+ "CREATE TABLE performance_reviews(review_id INTEGER, employee_id INTEGER"
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+ ", review_date DATE, rating INTEGER); "
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+ "INSERT INTO performance_reviews (review_id, employee_id, review_date, rating) VALUES"
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+ " (1, 1, '2023年01月15日', 2),"
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+ " (2, 1, '2023年04月15日', 3),"
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+ " (3, 1, '2023年07月15日', 4),"
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+ " (4, 1, '2023年10月15日', 5),"
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+ " (5, 2, '2023年02月01日', 3),"
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+ " (6, 2, '2023年05月01日', 2),"
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+ " (7, 2, '2023年08月01日', 4),"
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+ " (8, 2, '2023年11月01日', 5),"
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+ " (9, 3, '2023年03月10日', 1),"
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+ " (10, 3, '2023年06月10日', 2),"
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+ " (11, 3, '2023年09月10日', 3),"
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+ " (12, 3, '2023年12月10日', 4),"
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+ " (13, 4, '2023年01月20日', 4),"
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+ " (14, 4, '2023年04月20日', 4),"
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+ " (15, 4, '2023年07月20日', 4),"
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+ " (16, 5, '2023年02月15日', 3),"
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+ " (17, 5, '2023年05月15日', 2);")
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class MysqlTest {
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@Test
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void testScript(@EmbeddedDatabase DataSource dataSource)
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throws SQLException, FileNotFoundException {
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try (final Connection connection = dataSource.getConnection()) {
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try (final Statement statement = connection.createStatement();
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final ResultSet resultSet =
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statement.executeQuery(
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new BufferedReader(
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new FileReader(
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"src/main/java/g3501_3600/"
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+ "s3580_find_consistently_improving_employees/"
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+ "script.sql"))
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.lines()
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.collect(Collectors.joining("\n"))
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.replaceAll("#.*?\\r?\\n", ""))) {
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assertThat(resultSet.next(), equalTo(true));
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assertThat(resultSet.getNString(1), equalTo("2"));
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assertThat(resultSet.getNString(2), equalTo("Bob Smith"));
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assertThat(resultSet.getNString(3), equalTo("3"));
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assertThat(resultSet.next(), equalTo(true));
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assertThat(resultSet.getNString(1), equalTo("1"));
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assertThat(resultSet.getNString(2), equalTo("Alice Johnson"));
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assertThat(resultSet.getNString(3), equalTo("2"));
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assertThat(resultSet.next(), equalTo(true));
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assertThat(resultSet.getNString(1), equalTo("3"));
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assertThat(resultSet.getNString(2), equalTo("Carol Davis"));
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assertThat(resultSet.getNString(3), equalTo("2"));
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assertThat(resultSet.next(), equalTo(false));
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}
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}
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}
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}

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