Attempted 4 today

Author: mrinal1704

Easy/Delete duplicate emails.sql

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+-- Question 32
+-- Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.
+
+-- +----+------------------+
+-- | Id | Email            |
+-- +----+------------------+
+-- | 1  | [email protected] |
+-- | 2  | [email protected]  |
+-- | 3  | [email protected] |
+-- +----+------------------+
+-- Id is the primary key column for this table.
+-- For example, after running your query, the above Person table should have the following rows:
+
+-- +----+------------------+
+-- | Id | Email            |
+-- +----+------------------+
+-- | 1  | [email protected] |
+-- | 2  | [email protected]  |
+-- +----+------------------+
+
+-- Solution
+With t1 as
+(
+ Select *,
+    row_number() over(partition by email order by id) as rk
+    from person
+)
+Delete from person
+where id in (Select t1.id from t1 where t1.rk>1)

Easy/Sales Analysis 2.sql

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+-- Question 33
+-- Table: Product
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | product_id   | int     |
+-- | product_name | varchar |
+-- | unit_price   | int     |
+-- +--------------+---------+
+-- product_id is the primary key of this table.
+-- Table: Sales
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | seller_id   | int     |
+-- | product_id  | int     |
+-- | buyer_id    | int     |
+-- | sale_date   | date    |
+-- | quantity    | int     |
+-- | price       | int     |
+-- +------ ------+---------+
+-- This table has no primary key, it can have repeated rows.
+-- product_id is a foreign key to Product table.
+ 
+
+-- Write an SQL query that reports the buyers who have bought S8 but not iPhone. Note that S8 and iPhone are products present in the Product table.
+
+-- The query result format is in the following example:
+
+-- Product table:
+-- +------------+--------------+------------+
+-- | product_id | product_name | unit_price |
+-- +------------+--------------+------------+
+-- | 1          | S8           | 1000       |
+-- | 2          | G4           | 800        |
+-- | 3          | iPhone       | 1400       |
+-- +------------+--------------+------------+
+
+-- Sales table:
+-- +-----------+------------+----------+------------+----------+-------+
+-- | seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-- +-----------+------------+----------+------------+----------+-------+
+-- | 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
+-- | 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
+-- | 2         | 1          | 3        | 2019-06-02 | 1        | 800   |
+-- | 3         | 3          | 3        | 2019-05-13 | 2        | 2800  |
+-- +-----------+------------+----------+------------+----------+-------+
+
+-- Result table:
+-- +-------------+
+-- | buyer_id    |
+-- +-------------+
+-- | 1           |
+-- +-------------+
+-- The buyer with id 1 bought an S8 but didn't buy an iPhone. The buyer with id 3 bought both.
+
+-- Solution
+Select distinct a.buyer_id
+from sales a join
+product b
+on a.product_id = b.product_id
+where a.buyer_id in
+(Select a.buyer_id from sales a join product b on a.product_id = b.product_id where b.product_name = 'S8') 
+and
+a.buyer_id not in (Select a.buyer_id from sales a join product b on a.product_id = b.product_id where b.product_name = 'iPhone')

Easy/Sales Analysis 3.sql

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+-- Question 34
+-- Table: Product
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | product_id   | int     |
+-- | product_name | varchar |
+-- | unit_price   | int     |
+-- +--------------+---------+
+-- product_id is the primary key of this table.
+-- Table: Sales
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | seller_id   | int     |
+-- | product_id  | int     |
+-- | buyer_id    | int     |
+-- | sale_date   | date    |
+-- | quantity    | int     |
+-- | price       | int     |
+-- +------ ------+---------+
+-- This table has no primary key, it can have repeated rows.
+-- product_id is a foreign key to Product table.
+ 
+
+-- Write an SQL query that reports the products that were only sold in spring 2019. That is, between 2019-01-01 and 2019-03-31 inclusive.
+
+-- The query result format is in the following example:
+
+-- Product table:
+-- +------------+--------------+------------+
+-- | product_id | product_name | unit_price |
+-- +------------+--------------+------------+
+-- | 1          | S8           | 1000       |
+-- | 2          | G4           | 800        |
+-- | 3          | iPhone       | 1400       |
+-- +------------+--------------+------------+
+
+-- Sales table:
+-- +-----------+------------+----------+------------+----------+-------+
+-- | seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-- +-----------+------------+----------+------------+----------+-------+
+-- | 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
+-- | 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
+-- | 2         | 2          | 3        | 2019-06-02 | 1        | 800   |
+-- | 3         | 3          | 4        | 2019-05-13 | 2        | 2800  |
+-- +-----------+------------+----------+------------+----------+-------+
+
+-- Result table:
+-- +-------------+--------------+
+-- | product_id  | product_name |
+-- +-------------+--------------+
+-- | 1           | S8           |
+-- +-------------+--------------+
+-- The product with id 1 was only sold in spring 2019 while the other two were sold after.
+
+-- Solution
+select distinct a.product_id, product_name from sales a join product b on a.product_id = b.product_id where a.product_id 
+in
+(select product_id from sales where sale_date >= '2019-01-01' and sale_date <= '2019-03-31')
+and
+a.product_id not in 
+(select product_id from sales where sale_date > '2019-03-31' or sale_date < '2019-01-01')

Easy/User Activity for past 30 days 2.sql

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+-- Question 35
+-- 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 average number of sessions per user for a period of 30 days ending 2019-07-27 inclusively, rounded to 2 decimal places. The sessions we want to count for a user are those with at least one activity in that time period.
+
+-- 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   |
+-- | 3       | 5          | 2019-07-21    | open_session  |
+-- | 3       | 5          | 2019-07-21    | scroll_down   |
+-- | 3       | 5          | 2019-07-21    | end_session   |
+-- | 4       | 3          | 2019-06-25    | open_session  |
+-- | 4       | 3          | 2019-06-25    | end_session   |
+-- +---------+------------+---------------+---------------+
+
+-- Result table:
+-- +---------------------------+ 
+-- | average_sessions_per_user |
+-- +---------------------------+ 
+-- | 1.33                      |
+-- +---------------------------+ 
+-- User 1 and 2 each had 1 session in the past 30 days while user 3 had 2 sessions so the average is (1 + 1 + 2) / 3 = 1.33.
+
+
+-- Solution
+select ifnull(round(avg(a.num),2),0) as average_sessions_per_user
+from (
+select count(distinct session_id) as num
+from activity
+where activity_date between '2019-06-28' and '2019-07-27'
+group by user_id) a