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Medium/Article Views 2.sql

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+-- Question 81
+-- Table: Views
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | article_id    | int     |
+-- | author_id     | int     |
+-- | viewer_id     | int     |
+-- | view_date     | date    |
+-- +---------------+---------+
+-- There is no primary key for this table, it may have duplicate rows.
+-- Each row of this table indicates that some viewer viewed an article (written by some author) on some date. 
+-- Note that equal author_id and viewer_id indicate the same person.
+ 
+
+-- Write an SQL query to find all the people who viewed more than one article on the same date, sorted in ascending order by their id.
+
+-- The query result format is in the following example:
+
+-- Views table:
+-- +------------+-----------+-----------+------------+
+-- | article_id | author_id | viewer_id | view_date  |
+-- +------------+-----------+-----------+------------+
+-- | 1          | 3         | 5         | 2019-08-01 |
+-- | 3          | 4         | 5         | 2019-08-01 |
+-- | 1          | 3         | 6         | 2019-08-02 |
+-- | 2          | 7         | 7         | 2019-08-01 |
+-- | 2          | 7         | 6         | 2019-08-02 |
+-- | 4          | 7         | 1         | 2019-07-22 |
+-- | 3          | 4         | 4         | 2019-07-21 |
+-- | 3          | 4         | 4         | 2019-07-21 |
+-- +------------+-----------+-----------+------------+
+
+-- Result table:
+-- +------+
+-- | id   |
+-- +------+
+-- | 5    |
+-- | 6    |
+-- +------+
+
+-- Solution
+select distinct viewer_id as id#, count(distinct article_id) as total
+from views
+group by viewer_id, view_date
+having count(distinct article_id)>1
+order by 1

Medium/Immediate Food Delivery 2.sql

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+-- Question 82
+-- 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.
+
+-- The first order of a customer is the order with the earliest order date that customer made. It is guaranteed that a customer has exactly one first order.
+
+-- Write an SQL query to find the percentage of immediate orders in the first orders of all customers, 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           | 2           | 2019-08-02 | 2019-08-02                  |
+-- | 3           | 1           | 2019-08-11 | 2019-08-12                  |
+-- | 4           | 3           | 2019-08-24 | 2019-08-24                  |
+-- | 5           | 3           | 2019-08-21 | 2019-08-22                  |
+-- | 6           | 2           | 2019-08-11 | 2019-08-13                  |
+-- | 7           | 4           | 2019-08-09 | 2019-08-09                  |
+-- +-------------+-------------+------------+-----------------------------+
+
+-- Result table:
+-- +----------------------+
+-- | immediate_percentage |
+-- +----------------------+
+-- | 50.00                |
+-- +----------------------+
+-- The customer id 1 has a first order with delivery id 1 and it is scheduled.
+-- The customer id 2 has a first order with delivery id 2 and it is immediate.
+-- The customer id 3 has a first order with delivery id 5 and it is scheduled.
+-- The customer id 4 has a first order with delivery id 7 and it is immediate.
+-- Hence, half the customers have immediate first orders.
+
+-- Solution
+select 
+round(avg(case when order_date = customer_pref_delivery_date then 1 else 0 end)*100,2) as
+immediate_percentage
+from 
+(select *,
+ rank() over(partition by customer_id order by order_date) as rk
+from delivery) a
+where a.rk=1

Medium/Monthly Transactions 1.sql

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+-- Question 83
+-- Table: Transactions
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | country       | varchar |
+-- | state         | enum    |
+-- | amount        | int     |
+-- | trans_date    | date    |
+-- +---------------+---------+
+-- id is the primary key of this table.
+-- The table has information about incoming transactions.
+-- The state column is an enum of type ["approved", "declined"].
+ 
+
+-- Write an SQL query to find for each month and country, the number of transactions and their total amount, the number of approved transactions and their total amount.
+
+-- The query result format is in the following example:
+
+-- Transactions table:
+-- +------+---------+----------+--------+------------+
+-- | id   | country | state    | amount | trans_date |
+-- +------+---------+----------+--------+------------+
+-- | 121  | US      | approved | 1000   | 2018-12-18 |
+-- | 122  | US      | declined | 2000   | 2018-12-19 |
+-- | 123  | US      | approved | 2000   | 2019-01-01 |
+-- | 124  | DE      | approved | 2000   | 2019-01-07 |
+-- +------+---------+----------+--------+------------+
+
+-- Result table:
+-- +----------+---------+-------------+----------------+--------------------+-----------------------+
+-- | month    | country | trans_count | approved_count | trans_total_amount | approved_total_amount |
+-- +----------+---------+-------------+----------------+--------------------+-----------------------+
+-- | 2018-12  | US      | 2           | 1              | 3000               | 1000                  |
+-- | 2019-01  | US      | 1           | 1              | 2000               | 2000                  |
+-- | 2019-01  | DE      | 1           | 1              | 2000               | 2000                  |
+-- +----------+---------+-------------+----------------+--------------------+-----------------------+
+
+-- Solution
+with t1 as(
+select DATE_FORMAT(trans_date,'%Y-%m') as month, country, count(state) as trans_count, sum(amount) as trans_total_amount
+from transactions
+group by country, month(trans_date)),
+
+t2 as (
+Select DATE_FORMAT(trans_date,'%Y-%m') as month, country, count(state) as approved_count, sum(amount) as approved_total_amount
+from transactions
+where state = 'approved'
+group by country, month(trans_date))
+
+select t1.month, t1.country, coalesce(t1.trans_count,0) as trans_count, coalesce(t2.approved_count,0) as approved_count, coalesce(t1.trans_total_amount,0) as trans_total_amount, coalesce(t2.approved_total_amount,0) as approved_total_amount
+from t1 left join t2
+on t1.country = t2.country and t1.month = t2.month

Medium/Page Recommnedations.sql

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+-- Question 84
+-- Table: Friendship
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | user1_id      | int     |
+-- | user2_id      | int     |
+-- +---------------+---------+
+-- (user1_id, user2_id) is the primary key for this table.
+-- Each row of this table indicates that there is a friendship relation between user1_id and user2_id.
+ 
+
+-- Table: Likes
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | user_id     | int     |
+-- | page_id     | int     |
+-- +-------------+---------+
+-- (user_id, page_id) is the primary key for this table.
+-- Each row of this table indicates that user_id likes page_id.
+ 
+
+-- Write an SQL query to recommend pages to the user with user_id = 1 using the pages that your friends liked. It should not recommend pages you already liked.
+
+-- Return result table in any order without duplicates.
+
+-- The query result format is in the following example:
+
+-- Friendship table:
+-- +----------+----------+
+-- | user1_id | user2_id |
+-- +----------+----------+
+-- | 1        | 2        |
+-- | 1        | 3        |
+-- | 1        | 4        |
+-- | 2        | 3        |
+-- | 2        | 4        |
+-- | 2        | 5        |
+-- | 6        | 1        |
+-- +----------+----------+
+ 
+-- Likes table:
+-- +---------+---------+
+-- | user_id | page_id |
+-- +---------+---------+
+-- | 1       | 88      |
+-- | 2       | 23      |
+-- | 3       | 24      |
+-- | 4       | 56      |
+-- | 5       | 11      |
+-- | 6       | 33      |
+-- | 2       | 77      |
+-- | 3       | 77      |
+-- | 6       | 88      |
+-- +---------+---------+
+
+-- Result table:
+-- +------------------+
+-- | recommended_page |
+-- +------------------+
+-- | 23               |
+-- | 24               |
+-- | 56               |
+-- | 33               |
+-- | 77               |
+-- +------------------+
+-- User one is friend with users 2, 3, 4 and 6.
+-- Suggested pages are 23 from user 2, 24 from user 3, 56 from user 3 and 33 from user 6.
+-- Page 77 is suggested from both user 2 and user 3.
+-- Page 88 is not suggested because user 1 already likes it.
+
+-- Solution
+select distinct page_id as recommended_page
+from likes
+where user_id = 
+any(select user2_id as id
+from friendship
+where user1_id = 1 or user2_id = 1 and user2_id !=1
+union all
+select user1_id
+from friendship
+where user2_id = 1) 
+and page_id != all(select page_id from likes where user_id = 1)