Ezvor

3580. Find Consistently Improving Employees

MediumDatabase Preparing judge… LeetCode

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.

Database