solve: Medium faizanxmulla#13, #14

Author: faizanxmulla

interview-query-solutions/Medium/13-random-weighted-driver.sql

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+WITH cumulative_weights AS (
+    SELECT id,
+           SUM(weighting) OVER(ORDER BY id) AS cumulative_weight,
+           SUM(weighting) OVER() AS total_weight
+    FROM   drivers
+)
+SELECT   id
+FROM     cumulative_weights
+WHERE    cumulative_weight >= (
+            SELECT rand() * total_weight 
+            FROM   cumulative_weights 
+            LIMIT  1
+        )
+ORDER BY cumulative_weight
+LIMIT    1
+
+
+
+-- NOTE: 
+
+-- interesting and unique question. initially didnt have any idea how to even begin the question.
+-- use rand() in MySQL and random() in PostgresSQL.
+
+-- for detailed explanation, here is the ChatGPT chat link: 
+-- https://chatgpt.com/share/75b6430e-0120-49ec-aba2-be006f4d8d32

interview-query-solutions/Medium/14-employee-salaries.sql

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+-- my solution:
+
+WITH employees_cte as (
+    SELECT id
+           ,salary
+           ,department_id
+           ,COUNT(*) OVER(PARTITION BY department_id) as number_of_employees
+    FROM   employees
+)
+SELECT   DISTINCT name as department_name
+         ,number_of_employees
+         ,1.0 * COUNT(*) FILTER(WHERE salary > 100000) OVER(PARTITION BY department_id) / number_of_employees as percentage_over_100k
+FROM     employees_cte e JOIN departments d ON d.id=e.department_id
+WHERE    number_of_employees >= 10
+ORDER BY 3
+
+
+
+-- official solution:
+
+SELECT   d.name as department_name
+         ,AVG(CASE WHEN salary > 100000 THEN 1 ELSE 0 END) AS percentage_over_100k
+         ,COUNT(*) AS number_of_employees
+FROM     departments AS d JOIN employees AS e ON d.id = e.department_id
+GROUP BY 1
+HAVING   COUNT(*) >= 10
+ORDER BY 1 DESC
+LIMIT    3
+
+
+
+
+-- NOTE: 
+
+-- solved on own and in first attempt.
+-- seeing the solution realized that I did it in a way complicated manner.
+
+-- also, no need to use LEFT JOIN ; if the department has no employees, the condition ‘at least 10 employees’ won’t be satisfied.