task: #3580

Author: Ivanbyone

README.md

@@ -189,6 +189,7 @@ Useful for preparing for technical interviews and improving your SQL skills.
     - [3497. Analyze Subscription Conversion](./leetcode/medium/3497.%20Analyze%20Subscription%20Conversion.sql)
     - [3521. Find Product Recommendation Pairs](./leetcode/medium/3521.%20Find%20Product%20Recommendation%20Pairs.sql)
     - [3564. Seasonal Sales Analysis](./leetcode/medium/3564.%20Seasonal%20Sales%20Analysis.sql)
+    - [3580. Find Consistently Improving Employees](./leetcode/medium/3580.%20Find%20Consistently%20Improving%20Employees.sql)
     - [3601. Find Drivers with Improved Fuel Efficiency](./leetcode/medium/3601.%20Find%20Drivers%20with%20Improved%20Fuel%20Efficiency.sql)
 3. [Hard](./leetcode/hard/)
     - [185. Department Top Three Salaries](./leetcode/hard/185.%20Department%20Top%20Three%20Salaries.sql)

leetcode/medium/3580. Find Consistently Improving Employees.sql

@@ -0,0 +1,68 @@
+/*
+Question 3580. Find Consistently Improving Employees
+Link: https://leetcode.com/problems/find-consistently-improving-employees/description/?envType=problem-list-v2&envId=database
+
+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.
+*/
+
+WITH review_dates AS (
+    SELECT
+        employee_id,
+        MAX(review_date) AS review_date
+    FROM performance_reviews
+    GROUP BY employee_id
+),
+
+review_ratings AS (
+    SELECT 
+        pr.employee_id,
+        pr.review_date,
+        pr.rating,
+        LAG(pr.rating, 1, NULL) OVER (PARTITION BY pr.employee_id ORDER BY pr.review_date) AS second_rating,
+        LAG(pr.rating, 2, NULL) OVER (PARTITION BY pr.employee_id ORDER BY pr.review_date) AS third_rating
+    FROM performance_reviews AS pr
+)
+
+SELECT
+    e.employee_id,
+    e.name,
+    rr.rating - rr.third_rating AS improvement_score
+FROM employees AS e
+LEFT JOIN
+    review_dates AS rd
+    ON e.employee_id = rd.employee_id
+INNER JOIN
+    review_ratings AS rr
+    ON
+        e.employee_id = rr.employee_id
+        AND rd.review_date = rr.review_date
+WHERE rr.rating > rr.second_rating AND rr.second_rating > rr.third_rating
+ORDER BY improvement_score DESC, e.name ASC