Attempted 5

Author: mrinal1704

Medium/All people report to the given manager.sql

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+-- Question 55
+-- Table: Employees
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | employee_id   | int     |
+-- | employee_name | varchar |
+-- | manager_id    | int     |
+-- +---------------+---------+
+-- employee_id is the primary key for this table.
+-- Each row of this table indicates that the employee with ID employee_id and name employee_name reports his
+-- work to his/her direct manager with manager_id
+-- The head of the company is the employee with employee_id = 1.
+ 
+
+-- Write an SQL query to find employee_id of all employees that directly or indirectly report their work to the head of the company.
+
+-- The indirect relation between managers will not exceed 3 managers as the company is small.
+
+-- Return result table in any order without duplicates.
+
+-- The query result format is in the following example:
+
+-- Employees table:
+-- +-------------+---------------+------------+
+-- | employee_id | employee_name | manager_id |
+-- +-------------+---------------+------------+
+-- | 1           | Boss          | 1          |
+-- | 3           | Alice         | 3          |
+-- | 2           | Bob           | 1          |
+-- | 4           | Daniel        | 2          |
+-- | 7           | Luis          | 4          |
+-- | 8           | Jhon          | 3          |
+-- | 9           | Angela        | 8          |
+-- | 77          | Robert        | 1          |
+-- +-------------+---------------+------------+
+
+-- Result table:
+-- +-------------+
+-- | employee_id |
+-- +-------------+
+-- | 2           |
+-- | 77          |
+-- | 4           |
+-- | 7           |
+-- +-------------+
+
+-- The head of the company is the employee with employee_id 1.
+-- The employees with employee_id 2 and 77 report their work directly to the head of the company.
+-- The employee with employee_id 4 report his work indirectly to the head of the company 4 --> 2 --> 1. 
+-- The employee with employee_id 7 report his work indirectly to the head of the company 7 --> 4 --> 2 --> 1.
+-- The employees with employee_id 3, 8 and 9 don't report their work to head of company directly or indirectly.
+
+-- Solution
+select employee_id
+from employees
+where manager_id = 1 and employee_id != 1
+union
+select employee_id
+from employees
+where manager_id = any (select employee_id
+from employees
+where manager_id = 1 and employee_id != 1)
+union
+select employee_id
+from employees
+where manager_id = any (select employee_id
+from employees
+where manager_id = any (select employee_id
+from employees
+where manager_id = 1 and employee_id != 1))

Medium/Department Highest Salary.sql

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+-- Question 57
+-- The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id.
+
+-- +----+-------+--------+--------------+
+-- | Id | Name  | Salary | DepartmentId |
+-- +----+-------+--------+--------------+
+-- | 1  | Joe   | 70000  | 1            |
+-- | 2  | Jim   | 90000  | 1            |
+-- | 3  | Henry | 80000  | 2            |
+-- | 4  | Sam   | 60000  | 2            |
+-- | 5  | Max   | 90000  | 1            |
+-- +----+-------+--------+--------------+
+-- The Department table holds all departments of the company.
+
+-- +----+----------+
+-- | Id | Name     |
+-- +----+----------+
+-- | 1  | IT       |
+-- | 2  | Sales    |
+-- +----+----------+
+-- Write a SQL query to find employees who have the highest salary in each of the departments. 
+-- For the above tables, your SQL query should return the following rows (order of rows does not matter).
+
+-- +------------+----------+--------+
+-- | Department | Employee | Salary |
+-- +------------+----------+--------+
+-- | IT         | Max      | 90000  |
+-- | IT         | Jim      | 90000  |
+-- | Sales      | Henry    | 80000  |
+-- +------------+----------+--------+
+-- Explanation:
+
+-- Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.
+
+-- Solution
+select a.Department, a.Employee, a.Salary
+from(
+select d.name as Department, e.name as Employee, Salary,
+rank() over(partition by d.name order by salary desc) as rk
+from employee e
+join department d
+on e.departmentid = d.id) a
+where a.rk=1

Medium/Exchange Seats.sql

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+-- Question 56
+-- Mary is a teacher in a middle school and she has a table seat storing students' names and their corresponding seat ids.
+
+-- The column id is continuous increment.
+ 
+
+-- Mary wants to change seats for the adjacent students.
+ 
+
+-- Can you write a SQL query to output the result for Mary?
+ 
+
+-- +---------+---------+
+-- |    id   | student |
+-- +---------+---------+
+-- |    1    | Abbot   |
+-- |    2    | Doris   |
+-- |    3    | Emerson |
+-- |    4    | Green   |
+-- |    5    | Jeames  |
+-- +---------+---------+
+-- For the sample input, the output is:
+ 
+
+-- +---------+---------+
+-- |    id   | student |
+-- +---------+---------+
+-- |    1    | Doris   |
+-- |    2    | Abbot   |
+-- |    3    | Green   |
+-- |    4    | Emerson |
+-- |    5    | Jeames  |
+-- +---------+---------+
+
+-- Solution
+select row_number() over (order by (if(id%2=1,id+1,id-1))) as id, student
+from seat

Medium/NPV Queries.sql

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+-- Question 54
+-- Table: NPV
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | year          | int     |
+-- | npv           | int     |
+-- +---------------+---------+
+-- (id, year) is the primary key of this table.
+-- The table has information about the id and the year of each inventory and the corresponding net present value.
+ 
+
+-- Table: Queries
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | year          | int     |
+-- +---------------+---------+
+-- (id, year) is the primary key of this table.
+-- The table has information about the id and the year of each inventory query.
+ 
+
+-- Write an SQL query to find the npv of all each query of queries table.
+
+-- Return the result table in any order.
+
+-- The query result format is in the following example:
+
+-- NPV table:
+-- +------+--------+--------+
+-- | id   | year   | npv    |
+-- +------+--------+--------+
+-- | 1    | 2018   | 100    |
+-- | 7    | 2020   | 30     |
+-- | 13   | 2019   | 40     |
+-- | 1    | 2019   | 113    |
+-- | 2    | 2008   | 121    |
+-- | 3    | 2009   | 12     |
+-- | 11   | 2020   | 99     |
+-- | 7    | 2019   | 0      |
+-- +------+--------+--------+
+
+-- Queries table:
+-- +------+--------+
+-- | id   | year   |
+-- +------+--------+
+-- | 1    | 2019   |
+-- | 2    | 2008   |
+-- | 3    | 2009   |
+-- | 7    | 2018   |
+-- | 7    | 2019   |
+-- | 7    | 2020   |
+-- | 13   | 2019   |
+-- +------+--------+
+
+-- Result table:
+-- +------+--------+--------+
+-- | id   | year   | npv    |
+-- +------+--------+--------+
+-- | 1    | 2019   | 113    |
+-- | 2    | 2008   | 121    |
+-- | 3    | 2009   | 12     |
+-- | 7    | 2018   | 0      |
+-- | 7    | 2019   | 0      |
+-- | 7    | 2020   | 30     |
+-- | 13   | 2019   | 40     |
+-- +------+--------+--------+
+
+-- The npv value of (7, 2018) is not present in the NPV table, we consider it 0.
+-- The npv values of all other queries can be found in the NPV table.
+
+-- Solution
+select q.id, q.year, coalesce(n.npv,0) as npv
+from queries q
+left join npv n
+on q.id = n.id and q.year=n.year

Medium/Tree Node.sql

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+-- Question 58
+-- Given a table tree, id is identifier of the tree node and p_id is its parent node's id.
+
+-- +----+------+
+-- | id | p_id |
+-- +----+------+
+-- | 1  | null |
+-- | 2  | 1    |
+-- | 3  | 1    |
+-- | 4  | 2    |
+-- | 5  | 2    |
+-- +----+------+
+-- Each node in the tree can be one of three types:
+-- Leaf: if the node is a leaf node.
+-- Root: if the node is the root of the tree.
+-- Inner: If the node is neither a leaf node nor a root node.
+ 
+
+-- Write a query to print the node id and the type of the node. Sort your output by the node id. The result for the above sample is:
+ 
+
+-- +----+------+
+-- | id | Type |
+-- +----+------+
+-- | 1  | Root |
+-- | 2  | Inner|
+-- | 3  | Leaf |
+-- | 4  | Leaf |
+-- | 5  | Leaf |
+-- +----+------+
+ 
+
+-- Explanation
+
+ 
+
+-- Node '1' is root node, because its parent node is NULL and it has child node '2' and '3'.
+-- Node '2' is inner node, because it has parent node '1' and child node '4' and '5'.
+-- Node '3', '4' and '5' is Leaf node, because they have parent node and they don't have child node.
+
+-- And here is the image of the sample tree as below:
+ 
+
+-- 			  1
+-- 			/   \
+--           2       3
+--         /   \
+--       4       5
+-- Note
+
+-- If there is only one node on the tree, you only need to output its root attributes.
+
+-- Solution
+select id,
+case when p_id is null then 'Root'
+when id not in (select p_id from tree where p_id is not null group by p_id) then 'Leaf'
+else 'Inner'
+end as Type
+from tree
+order by id