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‎Medium/Calculate Salaries.sql‎

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-- Question 74
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-- Table Salaries:
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-- +---------------+---------+
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-- | Column Name | Type |
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-- +---------------+---------+
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-- | company_id | int |
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-- | employee_id | int |
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-- | employee_name | varchar |
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-- | salary | int |
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-- +---------------+---------+
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-- (company_id, employee_id) is the primary key for this table.
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-- This table contains the company id, the id, the name and the salary for an employee.
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-- Write an SQL query to find the salaries of the employees after applying taxes.
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-- The tax rate is calculated for each company based on the following criteria:
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-- 0% If the max salary of any employee in the company is less than 1000$.
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-- 24% If the max salary of any employee in the company is in the range [1000, 10000] inclusive.
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-- 49% If the max salary of any employee in the company is greater than 10000$.
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-- Return the result table in any order. Round the salary to the nearest integer.
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-- The query result format is in the following example:
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-- Salaries table:
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-- +------------+-------------+---------------+--------+
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-- | company_id | employee_id | employee_name | salary |
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-- +------------+-------------+---------------+--------+
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-- | 1 | 1 | Tony | 2000 |
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-- | 1 | 2 | Pronub | 21300 |
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-- | 1 | 3 | Tyrrox | 10800 |
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-- | 2 | 1 | Pam | 300 |
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-- | 2 | 7 | Bassem | 450 |
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-- | 2 | 9 | Hermione | 700 |
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-- | 3 | 7 | Bocaben | 100 |
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-- | 3 | 2 | Ognjen | 2200 |
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-- | 3 | 13 | Nyancat | 3300 |
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-- | 3 | 15 | Morninngcat | 1866 |
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-- +------------+-------------+---------------+--------+
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-- Result table:
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-- +------------+-------------+---------------+--------+
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-- | company_id | employee_id | employee_name | salary |
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-- +------------+-------------+---------------+--------+
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-- | 1 | 1 | Tony | 1020 |
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-- | 1 | 2 | Pronub | 10863 |
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-- | 1 | 3 | Tyrrox | 5508 |
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-- | 2 | 1 | Pam | 300 |
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-- | 2 | 7 | Bassem | 450 |
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-- | 2 | 9 | Hermione | 700 |
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-- | 3 | 7 | Bocaben | 76 |
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-- | 3 | 2 | Ognjen | 1672 |
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-- | 3 | 13 | Nyancat | 2508 |
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-- | 3 | 15 | Morninngcat | 5911 |
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-- +------------+-------------+---------------+--------+
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-- For company 1, Max salary is 21300. Employees in company 1 have taxes = 49%
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-- For company 2, Max salary is 700. Employees in company 2 have taxes = 0%
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-- For company 3, Max salary is 7777. Employees in company 3 have taxes = 24%
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-- The salary after taxes = salary - (taxes percentage / 100) * salary
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-- For example, Salary for Morninngcat (3, 15) after taxes = 7777 - 7777 * (24 / 100) = 7777 - 1866.48 = 5910.52, which is rounded to 5911.
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-- Solution
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with t1 as (
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select company_id, employee_id, employee_name, salary as sa, max(salary) over(partition by company_id) as maximum
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from salaries)
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select company_id, employee_id, employee_name,
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case when t1.maximum<1000 then t1.sa
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when t1.maximum between 1000 and 10000 then round(t1.sa*.76,0)
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else round(t1.sa*.51,0)
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end as salary
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from t1

‎Medium/Customers who bought a, b but not c.sql‎

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-- Question 72
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-- Table: Customers
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-- +---------------------+---------+
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-- | Column Name | Type |
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-- +---------------------+---------+
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-- | customer_id | int |
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-- | customer_name | varchar |
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-- +---------------------+---------+
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-- customer_id is the primary key for this table.
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-- customer_name is the name of the customer.
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-- Table: Orders
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-- +---------------+---------+
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-- | Column Name | Type |
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-- +---------------+---------+
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-- | order_id | int |
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-- | customer_id | int |
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-- | product_name | varchar |
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-- +---------------+---------+
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-- order_id is the primary key for this table.
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-- customer_id is the id of the customer who bought the product "product_name".
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-- Write an SQL query to report the customer_id and customer_name of customers who bought products "A", "B" but did not buy the product "C" since we want to recommend them buy this product.
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-- Return the result table ordered by customer_id.
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-- The query result format is in the following example.
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-- Customers table:
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-- +-------------+---------------+
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-- | customer_id | customer_name |
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-- +-------------+---------------+
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-- | 1 | Daniel |
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-- | 2 | Diana |
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-- | 3 | Elizabeth |
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-- | 4 | Jhon |
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-- +-------------+---------------+
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-- Orders table:
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-- +------------+--------------+---------------+
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-- | order_id | customer_id | product_name |
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-- +------------+--------------+---------------+
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-- | 10 | 1 | A |
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-- | 20 | 1 | B |
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-- | 30 | 1 | D |
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-- | 40 | 1 | C |
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-- | 50 | 2 | A |
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-- | 60 | 3 | A |
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-- | 70 | 3 | B |
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-- | 80 | 3 | D |
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-- | 90 | 4 | C |
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-- +------------+--------------+---------------+
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-- Result table:
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-- +-------------+---------------+
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-- | customer_id | customer_name |
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-- +-------------+---------------+
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-- | 3 | Elizabeth |
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-- +-------------+---------------+
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-- Only the customer_id with id 3 bought the product A and B but not the product C.
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-- Solution
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with t1 as
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(
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select customer_id
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from orders
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where product_name = 'B' and
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customer_id in (select customer_id
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from orders
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where product_name = 'A'))
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Select t1.customer_id, c.customer_name
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from t1 join customers c
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on t1.customer_id = c.customer_id
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where t1.customer_id != all(select customer_id
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from orders
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where product_name = 'C')

‎Medium/Managers with atleast 5 direct reports.sql‎

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-- Question 75
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-- The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
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-- +------+----------+-----------+----------+
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-- |Id |Name |Department |ManagerId |
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-- +------+----------+-----------+----------+
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-- |101 |John |A |null |
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-- |102 |Dan |A |101 |
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-- |103 |James |A |101 |
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-- |104 |Amy |A |101 |
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-- |105 |Anne |A |101 |
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-- |106 |Ron |B |101 |
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-- +------+----------+-----------+----------+
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-- Given the Employee table, write a SQL query that finds out managers with at least 5 direct report. For the above table, your SQL query should return:
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-- +-------+
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-- | Name |
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-- +-------+
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-- | John |
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-- +-------+
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-- Note:
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-- No one would report to himself.
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-- Solution
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with t1 as
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(
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select managerid, count(name) as total
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from employee
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group by managerid
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)
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select e.name
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from t1
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join employee e
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on t1.managerid = e.id
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where t1.total>=5

‎Medium/Reported Posts 2.sql‎

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-- Question 73
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-- Table: Actions
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-- +---------------+---------+
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-- | Column Name | Type |
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-- +---------------+---------+
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-- | user_id | int |
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-- | post_id | int |
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-- | action_date | date |
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-- | action | enum |
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-- | extra | varchar |
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-- +---------------+---------+
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-- There is no primary key for this table, it may have duplicate rows.
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-- The action column is an ENUM type of ('view', 'like', 'reaction', 'comment', 'report', 'share').
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-- The extra column has optional information about the action such as a reason for report or a type of reaction.
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-- Table: Removals
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-- +---------------+---------+
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-- | Column Name | Type |
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-- +---------------+---------+
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-- | post_id | int |
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-- | remove_date | date |
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-- +---------------+---------+
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-- post_id is the primary key of this table.
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-- Each row in this table indicates that some post was removed as a result of being reported or as a result of an admin review.
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-- Write an SQL query to find the average for daily percentage of posts that got removed after being reported as spam, rounded to 2 decimal places.
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-- The query result format is in the following example:
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-- Actions table:
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-- +---------+---------+-------------+--------+--------+
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-- | user_id | post_id | action_date | action | extra |
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-- +---------+---------+-------------+--------+--------+
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-- | 1 | 1 | 2019-07-01 | view | null |
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-- | 1 | 1 | 2019-07-01 | like | null |
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-- | 1 | 1 | 2019-07-01 | share | null |
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-- | 2 | 2 | 2019-07-04 | view | null |
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-- | 2 | 2 | 2019-07-04 | report | spam |
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-- | 3 | 4 | 2019-07-04 | view | null |
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-- | 3 | 4 | 2019-07-04 | report | spam |
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-- | 4 | 3 | 2019-07-02 | view | null |
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-- | 4 | 3 | 2019-07-02 | report | spam |
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-- | 5 | 2 | 2019-07-03 | view | null |
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-- | 5 | 2 | 2019-07-03 | report | racism |
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-- | 5 | 5 | 2019-07-03 | view | null |
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-- | 5 | 5 | 2019-07-03 | report | racism |
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-- +---------+---------+-------------+--------+--------+
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-- Removals table:
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-- +---------+-------------+
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-- | post_id | remove_date |
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-- +---------+-------------+
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-- | 2 | 2019-07-20 |
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-- | 3 | 2019-07-18 |
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-- +---------+-------------+
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-- Result table:
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-- +-----------------------+
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-- | average_daily_percent |
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-- +-----------------------+
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-- | 75.00 |
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-- +-----------------------+
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-- The percentage for 2019-07-04 is 50% because only one post of two spam reported posts was removed.
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-- The percentage for 2019-07-02 is 100% because one post was reported as spam and it was removed.
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-- The other days had no spam reports so the average is (50 + 100) / 2 = 75%
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-- Note that the output is only one number and that we do not care about the remove dates.
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-- Solution
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with t1 as(
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select a.action_date, (count(distinct r.post_id)+0.0)/(count(distinct a.post_id)+0.0) as result
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from (select action_date, post_id
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from actions
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where extra = 'spam' and action = 'report') a
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left join
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removals r
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on a.post_id = r.post_id
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group by a.action_date)
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select round(avg(t1.result)*100,2) as average_daily_percent
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from t1

‎Medium/Restaurant growth.sql‎

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-- Question 71
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-- Table: Customer
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-- +---------------+---------+
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-- | Column Name | Type |
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-- +---------------+---------+
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-- | customer_id | int |
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-- | name | varchar |
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-- | visited_on | date |
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-- | amount | int |
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-- +---------------+---------+
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-- (customer_id, visited_on) is the primary key for this table.
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-- This table contains data about customer transactions in a restaurant.
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-- visited_on is the date on which the customer with ID (customer_id) have visited the restaurant.
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-- amount is the total paid by a customer.
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-- You are the restaurant owner and you want to analyze a possible expansion (there will be at least one customer every day).
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-- Write an SQL query to compute moving average of how much customer paid in a 7 days window (current day + 6 days before) .
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-- The query result format is in the following example:
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-- Return result table ordered by visited_on.
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-- average_amount should be rounded to 2 decimal places, all dates are in the format ('YYYY-MM-DD').
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-- Customer table:
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-- +-------------+--------------+--------------+-------------+
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-- | customer_id | name | visited_on | amount |
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-- +-------------+--------------+--------------+-------------+
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-- | 1 | Jhon | 2019-01-01 | 100 |
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-- | 2 | Daniel | 2019-01-02 | 110 |
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-- | 3 | Jade | 2019-01-03 | 120 |
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-- | 4 | Khaled | 2019-01-04 | 130 |
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-- | 5 | Winston | 2019-01-05 | 110 |
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-- | 6 | Elvis | 2019-01-06 | 140 |
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-- | 7 | Anna | 2019-01-07 | 150 |
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-- | 8 | Maria | 2019-01-08 | 80 |
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-- | 9 | Jaze | 2019-01-09 | 110 |
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-- | 1 | Jhon | 2019-01-10 | 130 |
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-- | 3 | Jade | 2019-01-10 | 150 |
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-- +-------------+--------------+--------------+-------------+
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-- Result table:
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-- +--------------+--------------+----------------+
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-- | visited_on | amount | average_amount |
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-- +--------------+--------------+----------------+
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-- | 2019-01-07 | 860 | 122.86 |
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-- | 2019-01-08 | 840 | 120 |
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-- | 2019-01-09 | 840 | 120 |
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-- | 2019-01-10 | 1000 | 142.86 |
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-- +--------------+--------------+----------------+
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-- 1st moving average from 2019-01-01 to 2019-01-07 has an average_amount of (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
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-- 2nd moving average from 2019-01-02 to 2019-01-08 has an average_amount of (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
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-- 3rd moving average from 2019-01-03 to 2019-01-09 has an average_amount of (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
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-- 4th moving average from 2019-01-04 to 2019-01-10 has an average_amount of (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86
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-- Solution
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select visited_on, sum(amount) over(order by visited_on rows 6 preceding),
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round(avg(amount) over(order by visited_on rows 6 preceding),2)
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from
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(
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select visited_on, sum(amount) as amount
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from customer
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group by visited_on
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order by visited_on
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) a
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order by visited_on offset 6 rows

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