Table: employees +-------------+---------+ | Column Name | Type | +-------------+---------+ | employee_id | int | | name | varchar | +-------------+---------+ employee_id is the unique identifier for this table. Each row contains information about an employee. Table: performance_reviews +-------------+------+ | Column Name | Type | +-------------+------+ | review_id | int | | employee_id | int | | review_date | date | | rating | int | +-------------+------+ review_id is the unique identifier for this table. Each row represents a performance review for an employee. The rating is on a scale of 1-5 where 5 is excellent and 1 is poor. Write a solution to find employees who have consistently improved their performance over their last three reviews . An employee must have at least 3 review to be considered The employee's last 3 reviews must show strictly increasing ratings (each review better than the previous) Use the most recent 3 reviews based on review_date for each employee Calculate the improvement score as the difference between the latest rating and the earliest rating among the last 3 reviews Return the result table ordered by improvement score in descending order, then by name in ascending order . The result format is in the following example. Example: Input: employees table: +-------------+----------------+ | employee_id | name | +-------------+----------------+ | 1 | Alice Johnson | | 2 | Bob Smith | | 3 | Carol Davis | | 4 | David Wilson | | 5 | Emma Brown | +-------------+----------------+ performance_reviews table: +-----------+-------------+-------------+--------+ | review_id | employee_id | review_date | rating | +-----------+-------------+-------------+--------+ | 1 | 1 | 2023-01-15 | 2 | | 2 | 1 | 2023-04-15 | 3 | | 3 | 1 | 2023-07-15 | 4 | | 4 | 1 | 2023-10-15 | 5 | | 5 | 2 | 2023-02-01 | 3 | | 6 | 2 | 2023-05-01 | 2 | | 7 | 2 | 2023-08-01 | 4 | | 8 | 2 | 2023-11-01 | 5 | | 9 | 3 | 2023-03-10 | 1 | | 10 | 3 | 2023-06-10 | 2 | | 11 | 3 | 2023-09-10 | 3 | | 12 | 3 | 2023-12-10 | 4 | | 13 | 4 | 2023-01-20 | 4 | | 14 | 4 | 2023-04-20 | 4 | | 15 | 4 | 2023-07-20 | 4 | | 16 | 5 | 2023-02-15 | 3 | | 17 | 5 | 2023-05-15 | 2 | +-----------+-------------+-------------+--------+ Output: +-------------+----------------+-------------------+ | employee_id | name | improvement_score | +-------------+----------------+-------------------+ | 2 | Bob Smith | 3 | | 1 | Alice Johnson | 2 | | 3 | Carol Davis | 2 | +-------------+----------------+-------------------+ Explanation: Alice Johnson (employee_id = 1): Has 4 reviews with ratings: 2, 3, 4, 5 Last 3 reviews (by date): 2023-04-15 (3), 2023-07-15 (4), 2023-10-15 (5) Ratings are strictly increasing: 3 → 4 → 5 Improvement score: 5 - 3 = 2 Carol Davis (employee_id = 3): Has 4 reviews with ratings: 1, 2, 3, 4 Last 3 reviews (by date): 2023-06-10 (2), 2023-09-10 (3), 2023-12-10 (4) Ratings are strictly increasing: 2 → 3 → 4 Improvement score: 4 - 2 = 2 Bob Smith (employee_id = 2): Has 4 reviews with ratings: 3, 2, 4, 5 Last 3 reviews (by date): 2023-05-01 (2), 2023-08-01 (4), 2023-11-01 (5) Ratings are strictly increasing: 2 → 4 → 5 Improvement score: 5 - 2 = 3 Employees not included: David Wilson (employee_id = 4): Last 3 reviews are all 4 (no improvement) Emma Brown (employee_id = 5): Only has 2 reviews (needs at least 3) The output table is ordered by improvement_score in descending order, then by name in ascending order.