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‎README.md

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| # | Title | Difficulty | Tag | Time, ms | Time, %
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|------|----------------|-------------|-------------|----------|--------
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| 3580 |[Find Consistently Improving Employees](src/main/kotlin/g3501_3600/s3580_find_consistently_improving_employees)| Medium | Database | 449 | 91.67
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| 3579 |[Minimum Steps to Convert String with Operations](src/main/kotlin/g3501_3600/s3579_minimum_steps_to_convert_string_with_operations)| Hard | String, Dynamic_Programming, Greedy | 107 | 100.00
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| 3578 |[Count Partitions With Max-Min Difference at Most K](src/main/kotlin/g3501_3600/s3578_count_partitions_with_max_min_difference_at_most_k)| Medium | Array, Dynamic_Programming, Prefix_Sum, Sliding_Window, Queue, Monotonic_Queue | 33 | 100.00
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| 3577 |[Count the Number of Computer Unlocking Permutations](src/main/kotlin/g3501_3600/s3577_count_the_number_of_computer_unlocking_permutations)| Medium | Array, Math, Combinatorics, Brainteaser | 2 | 100.00

‎src/main/kotlin/g3501_3600/s3580_find_consistently_improving_employees/readme.md

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[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
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[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
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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 | 2023-01-15 | 2 |
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| 2 | 1 | 2023-04-15 | 3 |
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| 3 | 1 | 2023-07-15 | 4 |
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| 4 | 1 | 2023-10-15 | 5 |
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| 5 | 2 | 2023-02-01 | 3 |
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| 6 | 2 | 2023-05-01 | 2 |
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| 7 | 2 | 2023-08-01 | 4 |
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| 8 | 2 | 2023-11-01 | 5 |
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| 9 | 3 | 2023-03-10 | 1 |
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| 10 | 3 | 2023-06-10 | 2 |
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| 11 | 3 | 2023-09-10 | 3 |
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| 12 | 3 | 2023-12-10 | 4 |
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| 13 | 4 | 2023-01-20 | 4 |
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| 14 | 4 | 2023-04-20 | 4 |
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| 15 | 4 | 2023-07-20 | 4 |
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| 16 | 5 | 2023-02-15 | 3 |
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| 17 | 5 | 2023-05-15 | 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): 2023-04-15 (3), 2023-07-15 (4), 2023-10-15 (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): 2023-06-10 (2), 2023-09-10 (3), 2023-12-10 (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): 2023-05-01 (2), 2023-08-01 (4), 2023-11-01 (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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## Solution
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```sql
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# Write your MySQL query statement below
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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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```

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