Attempted 5 more today

Author: mrinal1704

Easy/Actors who cooperated with Directors atleast three times.sql

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+-- Question 21
+-- Table: ActorDirector
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | actor_id    | int     |
+-- | director_id | int     |
+-- | timestamp   | int     |
+-- +-------------+---------+
+-- timestamp is the primary key column for this table.
+ 
+
+-- Write a SQL query for a report that provides the pairs (actor_id, director_id) where the actor have cooperated with the director at least 3 times.
+
+-- Example:
+
+-- ActorDirector table:
+-- +-------------+-------------+-------------+
+-- | actor_id    | director_id | timestamp   |
+-- +-------------+-------------+-------------+
+-- | 1           | 1           | 0           |
+-- | 1           | 1           | 1           |
+-- | 1           | 1           | 2           |
+-- | 1           | 2           | 3           |
+-- | 1           | 2           | 4           |
+-- | 2           | 1           | 5           |
+-- | 2           | 1           | 6           |
+-- +-------------+-------------+-------------+
+
+-- Result table:
+-- +-------------+-------------+
+-- | actor_id    | director_id |
+-- +-------------+-------------+
+-- | 1           | 1           |
+-- +-------------+-------------+
+-- The only pair is (1, 1) where they cooperated exactly 3 times.
+
+-- Solution 
+Select actor_id, director_id
+from actordirector
+group by actor_id, director_id
+having count(*)>=3

Easy/Biggest Single number.sql

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+-- Question 24
+-- Table my_numbers contains many numbers in column num including duplicated ones.
+-- Can you write a SQL query to find the biggest number, which only appears once.
+
+-- +---+
+-- |num|
+-- +---+
+-- | 8 |
+-- | 8 |
+-- | 3 |
+-- | 3 |
+-- | 1 |
+-- | 4 |
+-- | 5 |
+-- | 6 | 
+-- For the sample data above, your query should return the following result:
+-- +---+
+-- |num|
+-- +---+
+-- | 6 |
+-- Note:
+-- If there is no such number, just output null.
+
+-- Solution
+Select max(a.num) as num
+from 
+( 
+    select num, count(*)
+    from my_numbers
+    group by num
+    having count(*)=1
+) a

Easy/Shortest Distance.sql

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+-- Question 25
+-- Table point holds the x coordinate of some points on x-axis in a plane, which are all integers.
+ 
+
+-- Write a query to find the shortest distance between two points in these points.
+ 
+
+-- | x   |
+-- |-----|
+-- | -1  |
+-- | 0   |
+-- | 2   |
+ 
+
+-- The shortest distance is '1' obviously, which is from point '-1' to '0'. So the output is as below:
+ 
+
+-- | shortest|
+-- |---------|
+-- | 1       |
+ 
+
+-- Note: Every point is unique, which means there is no duplicates in table point
+
+-- Solution
+select min(abs(abs(a.x)-abs(a.next_closest))) as shortest
+from(
+select *,
+lead(x) over(order by x) as next_closest
+from point) a 

Easy/Students and Examinations.sql

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+-- Question 23
+-- Table: Students
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | student_id    | int     |
+-- | student_name  | varchar |
+-- +---------------+---------+
+-- student_id is the primary key for this table.
+-- Each row of this table contains the ID and the name of one student in the school.
+ 
+
+-- Table: Subjects
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | subject_name | varchar |
+-- +--------------+---------+
+-- subject_name is the primary key for this table.
+-- Each row of this table contains the name of one subject in the school.
+ 
+
+-- Table: Examinations
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | student_id   | int     |
+-- | subject_name | varchar |
+-- +--------------+---------+
+-- There is no primary key for this table. It may contain duplicates.
+-- Each student from the Students table takes every course from Subjects table.
+-- Each row of this table indicates that a student with ID student_id attended the exam of subject_name.
+ 
+
+-- Write an SQL query to find the number of times each student attended each exam.
+
+-- Order the result table by student_id and subject_name.
+
+-- The query result format is in the following example:
+
+-- Students table:
+-- +------------+--------------+
+-- | student_id | student_name |
+-- +------------+--------------+
+-- | 1          | Alice        |
+-- | 2          | Bob          |
+-- | 13         | John         |
+-- | 6          | Alex         |
+-- +------------+--------------+
+-- Subjects table:
+-- +--------------+
+-- | subject_name |
+-- +--------------+
+-- | Math         |
+-- | Physics      |
+-- | Programming  |
+-- +--------------+
+-- Examinations table:
+-- +------------+--------------+
+-- | student_id | subject_name |
+-- +------------+--------------+
+-- | 1          | Math         |
+-- | 1          | Physics      |
+-- | 1          | Programming  |
+-- | 2          | Programming  |
+-- | 1          | Physics      |
+-- | 1          | Math         |
+-- | 13         | Math         |
+-- | 13         | Programming  |
+-- | 13         | Physics      |
+-- | 2          | Math         |
+-- | 1          | Math         |
+-- +------------+--------------+
+-- Result table:
+-- +------------+--------------+--------------+----------------+
+-- | student_id | student_name | subject_name | attended_exams |
+-- +------------+--------------+--------------+----------------+
+-- | 1          | Alice        | Math         | 3              |
+-- | 1          | Alice        | Physics      | 2              |
+-- | 1          | Alice        | Programming  | 1              |
+-- | 2          | Bob          | Math         | 1              |
+-- | 2          | Bob          | Physics      | 0              |
+-- | 2          | Bob          | Programming  | 1              |
+-- | 6          | Alex         | Math         | 0              |
+-- | 6          | Alex         | Physics      | 0              |
+-- | 6          | Alex         | Programming  | 0              |
+-- | 13         | John         | Math         | 1              |
+-- | 13         | John         | Physics      | 1              |
+-- | 13         | John         | Programming  | 1              |
+-- +------------+--------------+--------------+----------------+
+-- The result table should contain all students and all subjects.
+-- Alice attended Math exam 3 times, Physics exam 2 times and Programming exam 1 time.
+-- Bob attended Math exam 1 time, Programming exam 1 time and didn't attend the Physics exam.
+-- Alex didn't attend any exam.
+-- John attended Math exam 1 time, Physics exam 1 time and Programming exam 1 time.
+
+-- Solution
+Select a.student_id as student_id, a.student_name as student_name, a.subject_name as subject_name, coalesce(attended_exams,0) as attended_exams
+from(
+select *
+from students
+cross join subjects
+group by student_id, student_name, subject_name) a
+left join 
+(Select e.student_id, student_name, subject_name, count(*) as attended_exams
+from examinations e join students s
+on e.student_id = s.student_id
+group by e.student_id, student_name, subject_name) b
+on a.student_id = b.student_id and a.subject_name =b.subject_name
+order by a.student_id asc, a.subject_name asc

Easy/Swap Salary.sql

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+-- Question 22
+-- Given a table salary, such as the one below, that has m=male and f=female values. 
+-- Swap all f and m values (i.e., change all f values to m and vice versa) with 
+-- a single update statement and no intermediate temp table.
+
+-- Note that you must write a single update statement, DO NOT write any select statement for this problem.
+
+
+-- Example:
+
+-- | id | name | sex | salary |
+-- |----|------|-----|--------|
+-- | 1  | A    | m   | 2500   |
+-- | 2  | B    | f   | 1500   |
+-- | 3  | C    | m   | 5500   |
+-- | 4  | D    | f   | 500    |
+-- After running your update statement, the above salary table should have the following rows:
+-- | id | name | sex | salary |
+-- |----|------|-----|--------|
+-- | 1  | A    | f   | 2500   |
+-- | 2  | B    | m   | 1500   |
+-- | 3  | C    | f   | 5500   |
+-- | 4  | D    | m   | 500    |
+
+-- Solution
+Update salary
+set sex = Case when sex = 'm' then 'f' 
+when sex = 'f' then 'm'
+end;