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Author: mrinal1704

Medium/Count student number in departments.sql

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+-- Question 87
+-- A university uses 2 data tables, student and department, to store data about its students
+-- and the departments associated with each major.
+
+-- Write a query to print the respective department name and number of students majoring in each
+-- department for all departments in the department table (even ones with no current students).
+
+-- Sort your results by descending number of students; if two or more departments have the same number of students, 
+-- then sort those departments alphabetically by department name.
+
+-- The student is described as follow:
+
+-- | Column Name  | Type      |
+-- |--------------|-----------|
+-- | student_id   | Integer   |
+-- | student_name | String    |
+-- | gender       | Character |
+-- | dept_id      | Integer   |
+-- where student_id is the student's ID number, student_name is the student's name, gender is their gender, and dept_id is the department ID associated with their declared major.
+
+-- And the department table is described as below:
+
+-- | Column Name | Type    |
+-- |-------------|---------|
+-- | dept_id     | Integer |
+-- | dept_name   | String  |
+-- where dept_id is the department's ID number and dept_name is the department name.
+
+-- Here is an example input:
+-- student table:
+
+-- | student_id | student_name | gender | dept_id |
+-- |------------|--------------|--------|---------|
+-- | 1          | Jack         | M      | 1       |
+-- | 2          | Jane         | F      | 1       |
+-- | 3          | Mark         | M      | 2       |
+-- department table:
+
+-- | dept_id | dept_name   |
+-- |---------|-------------|
+-- | 1       | Engineering |
+-- | 2       | Science     |
+-- | 3       | Law         |
+-- The Output should be:
+
+-- | dept_name   | student_number |
+-- |-------------|----------------|
+-- | Engineering | 2              |
+-- | Science     | 1              |
+-- | Law         | 0              |
+
+-- Solution
+select dept_name, count(s.dept_id) as student_number
+from department d
+left join student s
+on d.dept_id = s.dept_id
+group by d.dept_id
+order by count(s.dept_id) desc, dept_name

Medium/Get highest answer rate question.sql

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+-- Question 86
+-- Get the highest answer rate question from a table survey_log with these columns: id, action, question_id, answer_id, q_num, timestamp.
+
+-- id means user id; action has these kind of values: "show", "answer", "skip"; answer_id is not null when action column is "answer", 
+-- while is null for "show" and "skip"; q_num is the numeral order of the question in current session.
+
+-- Write a sql query to identify the question which has the highest answer rate.
+
+-- Example:
+
+-- Input:
+-- +------+-----------+--------------+------------+-----------+------------+
+-- | id   | action    | question_id  | answer_id  | q_num     | timestamp  |
+-- +------+-----------+--------------+------------+-----------+------------+
+-- | 5    | show      | 285          | null       | 1         | 123        |
+-- | 5    | answer    | 285          | 124124     | 1         | 124        |
+-- | 5    | show      | 369          | null       | 2         | 125        |
+-- | 5    | skip      | 369          | null       | 2         | 126        |
+-- +------+-----------+--------------+------------+-----------+------------+
+-- Output:
+-- +-------------+
+-- | survey_log  |
+-- +-------------+
+-- |    285      |
+-- +-------------+
+-- Explanation:
+-- question 285 has answer rate 1/1, while question 369 has 0/1 answer rate, so output 285.
+ 
+
+-- Note: The highest answer rate meaning is: answer number's ratio in show number in the same question.
+
+-- Solution
+with t1 as(
+select a.question_id, coalesce(b.answer/a.show_1,0) as rate
+from 
+(select question_id, coalesce(count(*),0) as show_1
+from survey_log
+where action != 'answer'
+group by question_id) a
+left join
+(select question_id, coalesce(count(*),0) as answer
+from survey_log
+where action = 'answer'
+group by question_id) b
+on a.question_id = b.question_id)
+
+select a.question_id as survey_log
+from 
+( select t1.question_id,
+rank() over(order by rate desc) as rk
+from t1) a
+where a.rk = 1

Medium/Product Sales Analysis 3.sql

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+-- Question 90
+-- 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 selects the product id, year, quantity, and price for the first year of every product sold.
+
+-- 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 | first_year | quantity | price |
+-- +------------+------------+----------+-------+ 
+-- | 100        | 2008       | 10       | 5000  |
+-- | 200        | 2011       | 15       | 9000  |
+-- +------------+------------+----------+-------+
+
+-- Solution
+select a.product_id, a.year as first_year, a.quantity, a.price
+from
+( select product_id, quantity, price, year,
+  rank() over(partition by product_id order by year) as rk
+  from sales
+) a
+where a.rk = 1

Medium/Shortest distance in a plane.sql

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+-- Question 89
+-- Table point_2d holds the coordinates (x,y) of some unique points (more than two) in a plane.
+ 
+
+-- Write a query to find the shortest distance between these points rounded to 2 decimals.
+ 
+
+-- | x  | y  |
+-- |----|----|
+-- | -1 | -1 |
+-- | 0  | 0  |
+-- | -1 | -2 |
+ 
+
+-- The shortest distance is 1.00 from point (-1,-1) to (-1,2). So the output should be:
+ 
+
+-- | shortest |
+-- |----------|
+-- | 1.00     |
+ 
+
+-- Note: The longest distance among all the points are less than 10000.
+
+-- Solution
+select round(a.shortest,2) as shortest
+from(
+select sqrt(pow((p1.x-p2.x),2)+pow((p1.y-p2.y),2)) as shortest
+from point_2d p1
+cross join point_2d p2
+where p1.x!=p2.x or p1.y!=p2.y
+order by sqrt(pow((p1.x-p2.x),2)+pow((p1.y-p2.y),2))
+limit 1) a

Medium/Winning Candidate.sql

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+-- Question 88
+-- Table: Candidate
+
+-- +-----+---------+
+-- | id  | Name    |
+-- +-----+---------+
+-- | 1   | A       |
+-- | 2   | B       |
+-- | 3   | C       |
+-- | 4   | D       |
+-- | 5   | E       |
+-- +-----+---------+  
+-- Table: Vote
+
+-- +-----+--------------+
+-- | id  | CandidateId  |
+-- +-----+--------------+
+-- | 1   |     2        |
+-- | 2   |     4        |
+-- | 3   |     3        |
+-- | 4   |     2        |
+-- | 5   |     5        |
+-- +-----+--------------+
+-- id is the auto-increment primary key,
+-- CandidateId is the id appeared in Candidate table.
+-- Write a sql to find the name of the winning candidate, the above example will return the winner B.
+
+-- +------+
+-- | Name |
+-- +------+
+-- | B    |
+-- +------+
+-- Notes:
+
+-- You may assume there is no tie, in other words there will be only one winning candidate
+
+-- Solution
+with t1 as (
+select *, rank() over(order by b.votes desc) as rk
+from candidate c
+join 
+(select candidateid, count(*) as votes
+from vote
+group by candidateid) b
+on c.id = b.candidateid)
+
+select t1.name
+from t1
+where t1.rk=1