This Ruby and SQL Online Test
Separates Good From Bad Hires

Implement the unique_names method. When passed two arrays of names, it will return an array containing the names that appear in either or both arrays. The returned array should have no duplicates.

For example, calling unique_names(["Ava", "Emma", "Olivia"], ["Olivia", "Sophia", "Emma"]) should return an array containing Ava, Emma, Olivia, and Sophia in any order.

App usage data are kept in the following table:

TABLE sessions
  id INTEGER PRIMARY KEY,
  userId INTEGER NOT NULL,
  duration DECIMAL NOT NULL

Write a query that selects userId and average session duration for each user who has more than one session.

See the example case for more details.

The following data definition defines an organization's employee hierarchy.

An employee is a manager if any other employee has their managerId set to this employee's id. That means John is a manager if at least one other employee has their managerId set to John's id.

TABLE employees
  id INTEGER NOT NULL PRIMARY KEY
  managerId INTEGER
  name VARCHAR(30) NOT NULL
  FOREIGN KEY (managerId) REFERENCES employees(id)

Write a query that selects only the names of employees who are not managers.

See the example case for more details.

Given the following data definition, write a query that returns the number of students whose first name is John.

TABLE students
   id INTEGER PRIMARY KEY,
   firstName VARCHAR(30) NOT NULL,
   lastName VARCHAR(30) NOT NULL

A table containing the students enrolled in a yearly course has incorrect data in records with ids between 20 and 100 (inclusive).

TABLE enrollments
  id INTEGER NOT NULL PRIMARY KEY
  year INTEGER NOT NULL
  studentId INTEGER NOT NULL

Write a query that updates the field 'year' of every faulty record to 2015.

Information about pets is kept in two separate tables:

TABLE dogs
  id INTEGER NOT NULL PRIMARY KEY,
  name VARCHAR(50) NOT NULL

TABLE cats
  id INTEGER NOT NULL PRIMARY KEY,
  name VARCHAR(50) NOT NULL

Write a query that select all distinct pet names.

See the example case for more details.

A new social network site has the following data tables:

users
id	name	sex
1	Ann	null
2	Steve	m
3	Mary	f
4	Brenda	f

friends
user1	user2
1	2
1	3
2	3

Select data that will be returned by the following SQL query:

SELECT users.name, COUNT(*) as count FROM users
LEFT JOIN friends
ON users.id = friends.user1 OR users.id = friends.user2
WHERE users.sex = 'f'
GROUP BY users.id, users.name;

Each item in a web shop belongs to a seller. To ensure service quality, each seller has a rating.

The data are kept in the following two tables:

TABLE sellers
  id INTEGER PRIMARY KEY,
  name VARCHAR(30) NOT NULL,
  rating INTEGER NOT NULL

TABLE items
  id INTEGER PRIMARY KEY,
  name VARCHAR(30) NOT NULL,
  sellerId INTEGER
  FOREIGN KEY (sellerId) REFERENCES sellers(id)

Write a query that selects the item name and the name of its seller for each item that belongs to a seller with a rating greater than 4. The query should return the name of the item as the first column and name of the seller as the second column.

See the example case for more details.

Implement the function find_roots to find the roots of the quadratic equation: ax² + bx + c = 0. The function should return an array containing both roots in any order. If the equation has only one solution, the function should return that solution as both elements of the array. The equation will always have at least one solution.

The roots of the quadratic equation can be found with the following formula:

For example, find_roots(2, 10, 8) should return [-1, -4] or [-4, -1] as the roots of the equation 2x² + 10x + 8 = 0 are -1 and -4.

Implement a group_by_owners function that:

• Accepts a hash containing the file owner name for each file name.
• Returns a hash containing an array of file names for each owner name, in any order.

For example, for hash {'Input.txt' => 'Randy', 'Code.py' => 'Stan', 'Output.txt' => 'Randy'} the group_by_owners function should return {'Randy' => ['Input.txt', 'Output.txt'], 'Stan' => ['Code.py']}.

As part of a data processing pipeline, complete the implementation of the pipeline method:

• The method should accept a variable number of functions, and it should return a new function that accepts one parameter arg.
• The returned function should call the first function in the pipeline with the parameter arg, and call the second function with the result of the first function.
• The returned function should continue calling each function in the pipeline in order, following the same pattern, and return the value from the last function.

For example, pipeline(-> (x) { x * 3 }, -> (x) { x + 1 }, -> (x) { x / 2 }) then calling the returned function with 3 should return 5.

The following two tables are used to define users and their respective roles:

TABLE users
  id INTEGER NOT NULL PRIMARY KEY,
  userName VARCHAR(50) NOT NULL

TABLE roles
  id INTEGER NOT NULL PRIMARY KEY,
  role VARCHAR(20) NOT NULL

The users_roles table should contain the mapping between each user and their roles. Each user can have many roles, and each role can have many users.

Modify the provided SQL create table statement so that:

• Only users from the users table can exist within users_roles.
• Only roles from the roles table can exist within users_roles.
• A user can only have a specific role once.

See the example case for more details.

Implement the filter method. It should accept a list of students and a variable number of filters. The filters will be passed in as lambdas, each of which will take one student record and return a boolean value.

filter should return a list of student records that have passed through all the filters.

If no filters are provided, the list of student records should be returned as is. If no students pass through all the provided filters (or if no students are provided), return an empty array.

For example, consider the following code:

students = [
  { name: 'Thomas Edison', gpa: 3.45 },
  { name: 'Grace Hopper', gpa: 3.99 },
  { name: 'Leonardo DaVinci', gpa: 2.78 }
]

honor_roll = ->(record) { record[:gpa] > 3.0 }

filter(students, honor_roll) should return:

[{:name=>"Thomas Edison", :gpa=>3.45},{:name=>"Grace Hopper", :gpa=>3.99}]

An insurance company maintains records of sales made by its employees. Each employee is assigned to a state. States are grouped under regions. The following tables contain the data:

TABLE regions
  id INTEGER PRIMARY KEY
  name VARCHAR(50) NOT NULL

TABLE states
  id INTEGER PRIMARY KEY
  name VARCHAR(50) NOT NULL
  regionId INTEGER NOT NULL
  FOREIGN KEY (regionId) REFERENCES regions(id)

TABLE employees
  id INTEGER PRIMARY KEY
  name VARCHAR(50) NOT NULL
  stateId INTEGER NOT NULL
  FOREIGN KEY (stateId) REFERENCES states(id)

TABLE sales
  id INTEGER PRIMARY KEY
  amount INTEGER NOT NULL
  employeeId INTEGER NOT NULL
  FOREIGN KEY (employeeId) REFERENCES employees(id)  

Management requires a comparative region sales analysis report.

Write a query that returns:

• The region name.
• Average sales per employee for the region (Average sales = Total sales made for the region / Number of employees in the region).
• The difference between the average sales of the region with the highest average sales, and the average sales per employee for the region (average sales to be calculated as explained above).

Employees can have multiple sales. A region with no sales should be also returned. Use 0 for average sales per employee for such a region when calculating the 2nd and the 3rd column.

See the example case for more details.