Attempted 8 today

Author: mrinal1704

Easy/Article views.sql

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+-- Question 42
+-- 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 authors that viewed at least one of their own articles, 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 |
+-- | 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   |
+-- +------+
+-- | 4    |
+-- | 7    |
+-- +------+
+
+-- Solution
+select distinct author_id as id
+from views
+where author_id = viewer_id
+order by author_id

Easy/Find the team size.sql

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+-- Question 47
+-- Table: Employee
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | employee_id   | int     |
+-- | team_id       | int     |
+-- +---------------+---------+
+-- employee_id is the primary key for this table.
+-- Each row of this table contains the ID of each employee and their respective team.
+-- Write an SQL query to find the team size of each of the employees.
+
+-- Return result table in any order.
+
+-- The query result format is in the following example:
+
+-- Employee Table:
+-- +-------------+------------+
+-- | employee_id | team_id    |
+-- +-------------+------------+
+-- |     1       |     8      |
+-- |     2       |     8      |
+-- |     3       |     8      |
+-- |     4       |     7      |
+-- |     5       |     9      |
+-- |     6       |     9      |
+-- +-------------+------------+
+-- Result table:
+-- +-------------+------------+
+-- | employee_id | team_size  |
+-- +-------------+------------+
+-- |     1       |     3      |
+-- |     2       |     3      |
+-- |     3       |     3      |
+-- |     4       |     1      |
+-- |     5       |     2      |
+-- |     6       |     2      |
+-- +-------------+------------+
+-- Employees with Id 1,2,3 are part of a team with team_id = 8.
+-- Employees with Id 4 is part of a team with team_id = 7.
+-- Employees with Id 5,6 are part of a team with team_id = 9.
+
+-- Solution
+Select employee_id, b.team_size
+from employee e
+join 
+(
+Select team_id, count(team_id) as team_size
+from employee
+group by team_id) b
+on e.team_id = b.team_id

Easy/List the products ordered in a period.sql

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+-- Question 45
+-- Table: Products
+
+-- +------------------+---------+
+-- | Column Name      | Type    |
+-- +------------------+---------+
+-- | product_id       | int     |
+-- | product_name     | varchar |
+-- | product_category | varchar |
+-- +------------------+---------+
+-- product_id is the primary key for this table.
+-- This table contains data about the company's products.
+-- Table: Orders
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | product_id    | int     |
+-- | order_date    | date    |
+-- | unit          | int     |
+-- +---------------+---------+
+-- There is no primary key for this table. It may have duplicate rows.
+-- product_id is a foreign key to Products table.
+-- unit is the number of products ordered in order_date.
+ 
+
+-- Write an SQL query to get the names of products with greater than or equal to 100 units ordered in February 2020 and their amount.
+
+-- Return result table in any order.
+
+-- The query result format is in the following example:
+
+ 
+
+-- Products table:
+-- +-------------+-----------------------+------------------+
+-- | product_id  | product_name          | product_category |
+-- +-------------+-----------------------+------------------+
+-- | 1           | Leetcode Solutions    | Book             |
+-- | 2           | Jewels of Stringology | Book             |
+-- | 3           | HP                    | Laptop           |
+-- | 4           | Lenovo                | Laptop           |
+-- | 5           | Leetcode Kit          | T-shirt          |
+-- +-------------+-----------------------+------------------+
+
+-- Orders table:
+-- +--------------+--------------+----------+
+-- | product_id   | order_date   | unit     |
+-- +--------------+--------------+----------+
+-- | 1            | 2020-02-05   | 60       |
+-- | 1            | 2020-02-10   | 70       |
+-- | 2            | 2020-01-18   | 30       |
+-- | 2            | 2020-02-11   | 80       |
+-- | 3            | 2020-02-17   | 2        |
+-- | 3            | 2020-02-24   | 3        |
+-- | 4            | 2020-03-01   | 20       |
+-- | 4            | 2020-03-04   | 30       |
+-- | 4            | 2020-03-04   | 60       |
+-- | 5            | 2020-02-25   | 50       |
+-- | 5            | 2020-02-27   | 50       |
+-- | 5            | 2020-03-01   | 50       |
+-- +--------------+--------------+----------+
+
+-- Result table:
+-- +--------------------+---------+
+-- | product_name       | unit    |
+-- +--------------------+---------+
+-- | Leetcode Solutions | 130     |
+-- | Leetcode Kit       | 100     |
+-- +--------------------+---------+
+
+-- Products with product_id = 1 is ordered in February a total of (60 + 70) = 130.
+-- Products with product_id = 2 is ordered in February a total of 80.
+-- Products with product_id = 3 is ordered in February a total of (2 + 3) = 5.
+-- Products with product_id = 4 was not ordered in February 2020.
+-- Products with product_id = 5 is ordered in February a total of (50 + 50) = 100.
+
+-- Solution
+Select a.product_name, a.unit
+from
+(select p.product_name, sum(unit) as unit
+from orders o 
+join products p
+on o.product_id = p.product_id
+where month(order_date)=2 and year(order_date) = 2020
+group by o.product_id) a
+where a.unit>=100

Easy/Queries quality and percentage.sql

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+-- Question 41
+-- Table: Queries
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | query_name  | varchar |
+-- | result      | varchar |
+-- | position    | int     |
+-- | rating      | int     |
+-- +-------------+---------+
+-- There is no primary key for this table, it may have duplicate rows.
+-- This table contains information collected from some queries on a database.
+-- The position column has a value from 1 to 500.
+-- The rating column has a value from 1 to 5. Query with rating less than 3 is a poor query.
+ 
+
+-- We define query quality as:
+
+-- The average of the ratio between query rating and its position.
+
+-- We also define poor query percentage as:
+
+-- The percentage of all queries with rating less than 3.
+
+-- Write an SQL query to find each query_name, the quality and poor_query_percentage.
+
+-- Both quality and poor_query_percentage should be rounded to 2 decimal places.
+
+-- The query result format is in the following example:
+
+-- Queries table:
+-- +------------+-------------------+----------+--------+
+-- | query_name | result            | position | rating |
+-- +------------+-------------------+----------+--------+
+-- | Dog        | Golden Retriever  | 1        | 5      |
+-- | Dog        | German Shepherd   | 2        | 5      |
+-- | Dog        | Mule              | 200      | 1      |
+-- | Cat        | Shirazi           | 5        | 2      |
+-- | Cat        | Siamese           | 3        | 3      |
+-- | Cat        | Sphynx            | 7        | 4      |
+-- +------------+-------------------+----------+--------+
+
+-- Result table:
+-- +------------+---------+-----------------------+
+-- | query_name | quality | poor_query_percentage |
+-- +------------+---------+-----------------------+
+-- | Dog        | 2.50    | 33.33                 |
+-- | Cat        | 0.66    | 33.33                 |
+-- +------------+---------+-----------------------+
+
+-- Dog queries quality is ((5 / 1) + (5 / 2) + (1 / 200)) / 3 = 2.50
+-- Dog queries poor_ query_percentage is (1 / 3) * 100 = 33.33
+
+-- Cat queries quality equals ((2 / 5) + (3 / 3) + (4 / 7)) / 3 = 0.66
+-- Cat queries poor_ query_percentage is (1 / 3) * 100 = 33.33
+
+-- Solution
+Select query_name, round(sum(rating/position)/count(*),2) as quality, 
+round(avg(case when rating<3 then 1 else 0 end)*100,2) as poor_query_percentage
+from queries
+group by query_name

Easy/Reformat department table.sql

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+-- Question 44
+-- Table: Department
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | revenue       | int     |
+-- | month         | varchar |
+-- +---------------+---------+
+-- (id, month) is the primary key of this table.
+-- The table has information about the revenue of each department per month.
+-- The month has values in ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"].
+ 
+
+-- Write an SQL query to reformat the table such that there is a department id column and a revenue column for each month.
+
+-- The query result format is in the following example:
+
+-- Department table:
+-- +------+---------+-------+
+-- | id   | revenue | month |
+-- +------+---------+-------+
+-- | 1    | 8000    | Jan   |
+-- | 2    | 9000    | Jan   |
+-- | 3    | 10000   | Feb   |
+-- | 1    | 7000    | Feb   |
+-- | 1    | 6000    | Mar   |
+-- +------+---------+-------+
+
+-- Result table:
+-- +------+-------------+-------------+-------------+-----+-------------+
+-- | id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+-- +------+-------------+-------------+-------------+-----+-------------+
+-- | 1    | 8000        | 7000        | 6000        | ... | null        |
+-- | 2    | 9000        | null        | null        | ... | null        |
+-- | 3    | null        | 10000       | null        | ... | null        |
+-- +------+-------------+-------------+-------------+-----+-------------+
+
+-- Note that the result table has 13 columns (1 for the department id + 12 for the months).
+
+-- Solution
+select id,
+sum(if(month='Jan',revenue,null)) as Jan_Revenue,
+sum(if(month='Feb',revenue,null)) as Feb_Revenue,
+sum(if(month='Mar',revenue,null)) as Mar_Revenue,
+sum(if(month='Apr',revenue,null)) as Apr_Revenue,
+sum(if(month='May',revenue,null)) as May_Revenue,
+sum(if(month='Jun',revenue,null)) as Jun_Revenue,
+sum(if(month='Jul',revenue,null)) as Jul_Revenue,
+sum(if(month='Aug',revenue,null)) as Aug_Revenue,
+sum(if(month='Sep',revenue,null)) as Sep_Revenue,
+sum(if(month='Oct',revenue,null)) as Oct_Revenue,
+sum(if(month='Nov',revenue,null)) as Nov_Revenue,
+sum(if(month='Dec',revenue,null)) as Dec_Revenue
+from Department
+group by id

Easy/Replace employee id with unique identifier.sql

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+-- Question 48
+-- Table: Employees
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | name          | varchar |
+-- +---------------+---------+
+-- id is the primary key for this table.
+-- Each row of this table contains the id and the name of an employee in a company.
+ 
+
+-- Table: EmployeeUNI
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | unique_id     | int     |
+-- +---------------+---------+
+-- (id, unique_id) is the primary key for this table.
+-- Each row of this table contains the id and the corresponding unique id of an employee in the company.
+ 
+
+-- Write an SQL query to show the unique ID of each user, If a user doesn't have a unique ID replace just show null.
+
+-- Return the result table in any order.
+
+-- The query result format is in the following example:
+
+-- Employees table:
+-- +----+----------+
+-- | id | name     |
+-- +----+----------+
+-- | 1  | Alice    |
+-- | 7  | Bob      |
+-- | 11 | Meir     |
+-- | 90 | Winston  |
+-- | 3  | Jonathan |
+-- +----+----------+
+
+-- EmployeeUNI table:
+-- +----+-----------+
+-- | id | unique_id |
+-- +----+-----------+
+-- | 3  | 1         |
+-- | 11 | 2         |
+-- | 90 | 3         |
+-- +----+-----------+
+
+-- EmployeeUNI table:
+-- +-----------+----------+
+-- | unique_id | name     |
+-- +-----------+----------+
+-- | null      | Alice    |
+-- | null      | Bob      |
+-- | 2         | Meir     |
+-- | 3         | Winston  |
+-- | 1         | Jonathan |
+-- +-----------+----------+
+
+-- Alice and Bob don't have a unique ID, We will show null instead.
+-- The unique ID of Meir is 2.
+-- The unique ID of Winston is 3.
+-- The unique ID of Jonathan is 1.
+
+-- Solution
+select unique_id, name
+from employees e
+left join
+employeeuni u
+on e.id = u.id
+order by e.id