Attempted 6 today

Author: mrinal1704

Easy/Number of comments per post.sql

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+-- Question 31
+-- Table: Submissions
+
+-- +---------------+----------+
+-- | Column Name   | Type     |
+-- +---------------+----------+
+-- | sub_id        | int      |
+-- | parent_id     | int      |
+-- +---------------+----------+
+-- There is no primary key for this table, it may have duplicate rows.
+-- Each row can be a post or comment on the post.
+-- parent_id is null for posts.
+-- parent_id for comments is sub_id for another post in the table.
+ 
+
+-- Write an SQL query to find number of comments per each post.
+
+-- Result table should contain post_id and its corresponding number_of_comments, 
+-- and must be sorted by post_id in ascending order.
+
+-- Submissions may contain duplicate comments. You should count the number of unique comments per post.
+
+-- Submissions may contain duplicate posts. You should treat them as one post.
+
+-- The query result format is in the following example:
+
+-- Submissions table:
+-- +---------+------------+
+-- | sub_id  | parent_id  |
+-- +---------+------------+
+-- | 1       | Null       |
+-- | 2       | Null       |
+-- | 1       | Null       |
+-- | 12      | Null       |
+-- | 3       | 1          |
+-- | 5       | 2          |
+-- | 3       | 1          |
+-- | 4       | 1          |
+-- | 9       | 1          |
+-- | 10      | 2          |
+-- | 6       | 7          |
+-- +---------+------------+
+
+-- Result table:
+-- +---------+--------------------+
+-- | post_id | number_of_comments |
+-- +---------+--------------------+
+-- | 1       | 3                  |
+-- | 2       | 2                  |
+-- | 12      | 0                  |
+-- +---------+--------------------+
+
+-- The post with id 1 has three comments in the table with id 3, 4 and 9. The comment with id 3 is 
+-- repeated in the table, we counted it only once.
+-- The post with id 2 has two comments in the table with id 5 and 10.
+-- The post with id 12 has no comments in the table.
+-- The comment with id 6 is a comment on a deleted post with id 7 so we ignored it.
+
+-- Solution
+Select a.sub_id as post_id, coalesce(b.number_of_comments,0) as number_of_comments
+from(
+select distinct sub_id from submissions where parent_id is null) a
+left join(
+select parent_id, count(distinct(sub_id)) as number_of_comments
+from submissions
+group by parent_id
+having parent_id = any(select sub_id from submissions where parent_id is null)) b
+on a.sub_id = b.parent_id
+order by post_id

Easy/Product Sales Analysis 1.sql

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+-- Question 30
+-- Table: Sales
+
+-- +-------------+-------+
+-- | Column Name | Type  |
+-- +-------------+-------+
+-- | sale_id     | int   |
+-- | product_id  | int   |
+-- | year        | int   |
+-- | quantity    | int   |
+-- | price       | int   |
+-- +-------------+-------+
+-- (sale_id, year) is the primary key of this table.
+-- product_id is a foreign key to Product table.
+-- Note that the price is per unit.
+-- Table: Product
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | product_id   | int     |
+-- | product_name | varchar |
+-- +--------------+---------+
+-- product_id is the primary key of this table.
+ 
+
+-- Write an SQL query that reports all product names of the products in the Sales table along with their selling year and price.
+
+-- For example:
+
+-- Sales table:
+-- +---------+------------+------+----------+-------+
+-- | sale_id | product_id | year | quantity | price |
+-- +---------+------------+------+----------+-------+ 
+-- | 1       | 100        | 2008 | 10       | 5000  |
+-- | 2       | 100        | 2009 | 12       | 5000  |
+-- | 7       | 200        | 2011 | 15       | 9000  |
+-- +---------+------------+------+----------+-------+
+
+-- Product table:
+-- +------------+--------------+
+-- | product_id | product_name |
+-- +------------+--------------+
+-- | 100        | Nokia        |
+-- | 200        | Apple        |
+-- | 300        | Samsung      |
+-- +------------+--------------+
+
+-- Result table:
+-- +--------------+-------+-------+
+-- | product_name | year  | price |
+-- +--------------+-------+-------+
+-- | Nokia        | 2008  | 5000  |
+-- | Nokia        | 2009  | 5000  |
+-- | Apple        | 2011  | 9000  |
+-- +--------------+-------+-------+
+
+-- Solution
+Select a.product_name, b.year, b.price
+from product as a
+join
+sales as b
+on a.product_id = b.product_id

Easy/Product Sales Analysis 2.sql

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+-- Question 29
+-- Table: Sales
+
+-- +-------------+-------+
+-- | Column Name | Type  |
+-- +-------------+-------+
+-- | sale_id     | int   |
+-- | product_id  | int   |
+-- | year        | int   |
+-- | quantity    | int   |
+-- | price       | int   |
+-- +-------------+-------+
+-- sale_id is the primary key of this table.
+-- product_id is a foreign key to Product table.
+-- Note that the price is per unit.
+-- Table: Product
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | product_id   | int     |
+-- | product_name | varchar |
+-- +--------------+---------+
+-- product_id is the primary key of this table.
+ 
+
+-- Write an SQL query that reports the total quantity sold for every product id.
+
+-- The query result format is in the following example:
+
+-- Sales table:
+-- +---------+------------+------+----------+-------+
+-- | sale_id | product_id | year | quantity | price |
+-- +---------+------------+------+----------+-------+ 
+-- | 1       | 100        | 2008 | 10       | 5000  |
+-- | 2       | 100        | 2009 | 12       | 5000  |
+-- | 7       | 200        | 2011 | 15       | 9000  |
+-- +---------+------------+------+----------+-------+
+
+-- Product table:
+-- +------------+--------------+
+-- | product_id | product_name |
+-- +------------+--------------+
+-- | 100        | Nokia        |
+-- | 200        | Apple        |
+-- | 300        | Samsung      |
+-- +------------+--------------+
+
+-- Result table:
+-- +--------------+----------------+
+-- | product_id   | total_quantity |
+-- +--------------+----------------+
+-- | 100          | 22             |
+-- | 200          | 15             |
+-- +--------------+----------------+
+
+-- Solution
+Select a.product_id, sum(a.quantity) as total_quantity
+from sales a
+join
+product b
+on a.product_id = b.product_id
+group by a.product_id

Easy/Project Employees 1.sql

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+-- Question 26
+-- Table: Project
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | project_id  | int     |
+-- | employee_id | int     |
+-- +-------------+---------+
+-- (project_id, employee_id) is the primary key of this table.
+-- employee_id is a foreign key to Employee table.
+-- Table: Employee
+
+-- +------------------+---------+
+-- | Column Name      | Type    |
+-- +------------------+---------+
+-- | employee_id      | int     |
+-- | name             | varchar |
+-- | experience_years | int     |
+-- +------------------+---------+
+-- employee_id is the primary key of this table.
+ 
+
+-- Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.
+
+-- The query result format is in the following example:
+
+-- Project table:
+-- +-------------+-------------+
+-- | project_id  | employee_id |
+-- +-------------+-------------+
+-- | 1           | 1           |
+-- | 1           | 2           |
+-- | 1           | 3           |
+-- | 2           | 1           |
+-- | 2           | 4           |
+-- +-------------+-------------+
+
+-- Employee table:
+-- +-------------+--------+------------------+
+-- | employee_id | name   | experience_years |
+-- +-------------+--------+------------------+
+-- | 1           | Khaled | 3                |
+-- | 2           | Ali    | 2                |
+-- | 3           | John   | 1                |
+-- | 4           | Doe    | 2                |
+-- +-------------+--------+------------------+
+
+-- Result table:
+-- +-------------+---------------+
+-- | project_id  | average_years |
+-- +-------------+---------------+
+-- | 1           | 2.00          |
+-- | 2           | 2.50          |
+-- +-------------+---------------+
+-- The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50
+
+-- Solution
+Select a.project_id, round(sum(b.experience_years)/count(b.employee_id),2) as average_years
+from project as a
+join
+employee as b
+on a.employee_id=b.employee_id
+group by a.project_id

Easy/Project Employees 2.sql

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+-- Question 28
+-- Table: Project
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | project_id  | int     |
+-- | employee_id | int     |
+-- +-------------+---------+
+-- (project_id, employee_id) is the primary key of this table.
+-- employee_id is a foreign key to Employee table.
+-- Table: Employee
+
+-- +------------------+---------+
+-- | Column Name      | Type    |
+-- +------------------+---------+
+-- | employee_id      | int     |
+-- | name             | varchar |
+-- | experience_years | int     |
+-- +------------------+---------+
+-- employee_id is the primary key of this table.
+ 
+
+-- Write an SQL query that reports all the projects that have the most employees.
+
+-- The query result format is in the following example:
+
+-- Project table:
+-- +-------------+-------------+
+-- | project_id  | employee_id |
+-- +-------------+-------------+
+-- | 1           | 1           |
+-- | 1           | 2           |
+-- | 1           | 3           |
+-- | 2           | 1           |
+-- | 2           | 4           |
+-- +-------------+-------------+
+
+-- Employee table:
+-- +-------------+--------+------------------+
+-- | employee_id | name   | experience_years |
+-- +-------------+--------+------------------+
+-- | 1           | Khaled | 3                |
+-- | 2           | Ali    | 2                |
+-- | 3           | John   | 1                |
+-- | 4           | Doe    | 2                |
+-- +-------------+--------+------------------+
+
+-- Result table:
+-- +-------------+
+-- | project_id  |
+-- +-------------+
+-- | 1           |
+-- +-------------+
+-- The first project has 3 employees while the second one has 2.
+
+-- Solution
+select a.project_id
+from(
+select project_id,
+rank() over(order by count(employee_id) desc) as rk
+from project
+group by project_id) a
+where a.rk = 1