Attempted 5 more today

Author: mrinal1704

Easy/Average selling price.sql

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+-- Question 39
+-- Table: Prices
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | product_id    | int     |
+-- | start_date    | date    |
+-- | end_date      | date    |
+-- | price         | int     |
+-- +---------------+---------+
+-- (product_id, start_date, end_date) is the primary key for this table.
+-- Each row of this table indicates the price of the product_id in the period from start_date to end_date.
+-- For each product_id there will be no two overlapping periods. That means there will be no two intersecting periods for the same product_id.
+ 
+
+-- Table: UnitsSold
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | product_id    | int     |
+-- | purchase_date | date    |
+-- | units         | int     |
+-- +---------------+---------+
+-- There is no primary key for this table, it may contain duplicates.
+-- Each row of this table indicates the date, units and product_id of each product sold. 
+ 
+
+-- Write an SQL query to find the average selling price for each product.
+
+-- average_price should be rounded to 2 decimal places.
+
+-- The query result format is in the following example:
+
+-- Prices table:
+-- +------------+------------+------------+--------+
+-- | product_id | start_date | end_date   | price  |
+-- +------------+------------+------------+--------+
+-- | 1          | 2019-02-17 | 2019-02-28 | 5      |
+-- | 1          | 2019-03-01 | 2019-03-22 | 20     |
+-- | 2          | 2019-02-01 | 2019-02-20 | 15     |
+-- | 2          | 2019-02-21 | 2019-03-31 | 30     |
+-- +------------+------------+------------+--------+
+ 
+-- UnitsSold table:
+-- +------------+---------------+-------+
+-- | product_id | purchase_date | units |
+-- +------------+---------------+-------+
+-- | 1          | 2019-02-25    | 100   |
+-- | 1          | 2019-03-01    | 15    |
+-- | 2          | 2019-02-10    | 200   |
+-- | 2          | 2019-03-22    | 30    |
+-- +------------+---------------+-------+
+
+-- Result table:
+-- +------------+---------------+
+-- | product_id | average_price |
+-- +------------+---------------+
+-- | 1          | 6.96          |
+-- | 2          | 16.96         |
+-- +------------+---------------+
+-- Average selling price = Total Price of Product / Number of products sold.
+-- Average selling price for product 1 = ((100 * 5) + (15 * 20)) / 115 = 6.96
+-- Average selling price for product 2 = ((200 * 15) + (30 * 30)) / 230 = 16.96
+
+-- Solution
+Select d.product_id, round((sum(price*units)+0.00)/(sum(units)+0.00),2) as average_price
+from(
+Select *
+from prices p
+natural join 
+unitssold u
+where u.purchase_date between p.start_date and p.end_date) d
+group by d.product_id 

Easy/Consecutive available seats.sql

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+-- Question 37
+-- Several friends at a cinema ticket office would like to reserve consecutive available seats.
+-- Can you help to query all the consecutive available seats order by the seat_id using the following cinema table?
+-- | seat_id | free |
+-- |---------|------|
+-- | 1       | 1    |
+-- | 2       | 0    |
+-- | 3       | 1    |
+-- | 4       | 1    |
+-- | 5       | 1    |
+ 
+
+-- Your query should return the following result for the sample case above.
+ 
+
+-- | seat_id |
+-- |---------|
+-- | 3       |
+-- | 4       |
+-- | 5       |
+-- Note:
+-- The seat_id is an auto increment int, and free is bool ('1' means free, and '0' means occupied.).
+-- Consecutive available seats are more than 2(inclusive) seats consecutively available.
+
+-- Solution
+Select seat_id
+from(
+select seat_id, free,
+lead(free,1) over() as next,
+lag(free,1) over() as prev
+from cinema) a
+where a.free=True and (next = True or prev=True)
+order by seat_id

Easy/Immediate food delivery.sql

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+-- Question 38
+-- Table: Delivery
+
+-- +-----------------------------+---------+
+-- | Column Name                 | Type    |
+-- +-----------------------------+---------+
+-- | delivery_id                 | int     |
+-- | customer_id                 | int     |
+-- | order_date                  | date    |
+-- | customer_pref_delivery_date | date    |
+-- +-----------------------------+---------+
+-- delivery_id is the primary key of this table.
+-- The table holds information about food delivery to customers that make orders at some date and specify a preferred delivery date (on the same order date or after it).
+ 
+
+-- If the preferred delivery date of the customer is the same as the order date then the order is called immediate otherwise it's called scheduled.
+
+-- Write an SQL query to find the percentage of immediate orders in the table, rounded to 2 decimal places.
+
+-- The query result format is in the following example:
+
+-- Delivery table:
+-- +-------------+-------------+------------+-----------------------------+
+-- | delivery_id | customer_id | order_date | customer_pref_delivery_date |
+-- +-------------+-------------+------------+-----------------------------+
+-- | 1           | 1           | 2019-08-01 | 2019-08-02                  |
+-- | 2           | 5           | 2019-08-02 | 2019-08-02                  |
+-- | 3           | 1           | 2019-08-11 | 2019-08-11                  |
+-- | 4           | 3           | 2019-08-24 | 2019-08-26                  |
+-- | 5           | 4           | 2019-08-21 | 2019-08-22                  |
+-- | 6           | 2           | 2019-08-11 | 2019-08-13                  |
+-- +-------------+-------------+------------+-----------------------------+
+
+-- Result table:
+-- +----------------------+
+-- | immediate_percentage |
+-- +----------------------+
+-- | 33.33                |
+-- +----------------------+
+-- The orders with delivery id 2 and 3 are immediate while the others are scheduled.
+
+-- Solution
+Select 
+Round(avg(case when order_date=customer_pref_delivery_date then 1 else 0 end)*100,2) as immediate_percentage
+from delivery

Easy/Students with invalid departments.sql

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+-- Question 36
+-- Table: Departments
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | name          | varchar |
+-- +---------------+---------+
+-- id is the primary key of this table.
+-- The table has information about the id of each department of a university.
+ 
+
+-- Table: Students
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | name          | varchar |
+-- | department_id | int     |
+-- +---------------+---------+
+-- id is the primary key of this table.
+-- The table has information about the id of each student at a university and the id of the department he/she studies at.
+ 
+
+-- Write an SQL query to find the id and the name of all students who are enrolled in departments that no longer exists.
+
+-- Return the result table in any order.
+
+-- The query result format is in the following example:
+
+-- Departments table:
+-- +------+--------------------------+
+-- | id   | name                     |
+-- +------+--------------------------+
+-- | 1    | Electrical Engineering   |
+-- | 7    | Computer Engineering     |
+-- | 13   | Bussiness Administration |
+-- +------+--------------------------+
+
+-- Students table:
+-- +------+----------+---------------+
+-- | id   | name     | department_id |
+-- +------+----------+---------------+
+-- | 23   | Alice    | 1             |
+-- | 1    | Bob      | 7             |
+-- | 5    | Jennifer | 13            |
+-- | 2    | John     | 14            |
+-- | 4    | Jasmine  | 77            |
+-- | 3    | Steve    | 74            |
+-- | 6    | Luis     | 1             |
+-- | 8    | Jonathan | 7             |
+-- | 7    | Daiana   | 33            |
+-- | 11   | Madelynn | 1             |
+-- +------+----------+---------------+
+
+-- Result table:
+-- +------+----------+
+-- | id   | name     |
+-- +------+----------+
+-- | 2    | John     |
+-- | 7    | Daiana   |
+-- | 4    | Jasmine  |
+-- | 3    | Steve    |
+-- +------+----------+
+
+-- John, Daiana, Steve and Jasmine are enrolled in departments 14, 33, 74 and 77 respectively. 
+-- department 14, 33, 74 and 77 doesn't exist in the Departments table.
+
+-- Solution
+Select s.id, s.name
+from students s left join
+departments d
+on s.department_id = d.id
+where d.name is null

Easy/User activity for past 30 days 1.sql

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+-- Question 40
+-- Table: Activity
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | user_id       | int     |
+-- | session_id    | int     |
+-- | activity_date | date    |
+-- | activity_type | enum    |
+-- +---------------+---------+
+-- There is no primary key for this table, it may have duplicate rows.
+-- The activity_type column is an ENUM of type ('open_session', 'end_session', 'scroll_down', 'send_message').
+-- The table shows the user activities for a social media website. 
+-- Note that each session belongs to exactly one user.
+ 
+
+-- Write an SQL query to find the daily active user count for a period of 30 days ending 2019-07-27 inclusively. A user was active on some day if he/she made at least one activity on that day.
+
+-- The query result format is in the following example:
+
+-- Activity table:
+-- +---------+------------+---------------+---------------+
+-- | user_id | session_id | activity_date | activity_type |
+-- +---------+------------+---------------+---------------+
+-- | 1       | 1          | 2019-07-20    | open_session  |
+-- | 1       | 1          | 2019-07-20    | scroll_down   |
+-- | 1       | 1          | 2019-07-20    | end_session   |
+-- | 2       | 4          | 2019-07-20    | open_session  |
+-- | 2       | 4          | 2019-07-21    | send_message  |
+-- | 2       | 4          | 2019-07-21    | end_session   |
+-- | 3       | 2          | 2019-07-21    | open_session  |
+-- | 3       | 2          | 2019-07-21    | send_message  |
+-- | 3       | 2          | 2019-07-21    | end_session   |
+-- | 4       | 3          | 2019-06-25    | open_session  |
+-- | 4       | 3          | 2019-06-25    | end_session   |
+-- +---------+------------+---------------+---------------+
+
+-- Result table:
+-- +------------+--------------+ 
+-- | day        | active_users |
+-- +------------+--------------+ 
+-- | 2019-07-20 | 2            |
+-- | 2019-07-21 | 2            |
+-- +------------+--------------+ 
+-- Note that we do not care about days with zero active users.
+
+-- Solution
+Select activity_date as day, count(distinct user_id) as active_users
+from activity
+where activity_date > '2019-06-26' and activity_date < '2019-07-27'
+group by activity_date