This Python and SQL Online Test
Separates Good From Bad Hires

A university's Office of Admission keeps track of student majors. Each student can have a single major. Below is an example of how their system stores students, majors, and how it manipulates them:

students = [("Allen Anderson", "Computer Science"),
            ("Edgar Einstein", "Engineering"),
            ("Farrah Finn", "Fine Arts")]
     
def add_new_student(students, name, major):
    students.append((name, major))

def update_student(students, index, name, major):
    students[index] = (name, major)

def find_students_by_name(students, name):
    return [student for student in students if name in student[0]]

def get_all_majors(students):
    return [student[1] for student in students]

What can be concluded from the code snippet above?

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;

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.

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 a group_by_owners function that:

• Accepts a dictionary containing the file owner name for each file name.
• Returns a dictionary containing a list of file names for each owner name, in any order.

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

Implement the IceCreamMachine's scoops method so that it returns all combinations of one ingredient and one topping. If there are no ingredients or toppings, the method should return an empty list.

For example, IceCreamMachine(["vanilla", "chocolate"], ["chocolate sauce"]).scoops() should return [['vanilla', 'chocolate sauce'], ['chocolate', 'chocolate sauce']].

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

For example, calling unique_names(['Ava', 'Emma', 'Olivia'], ['Olivia', 'Sophia', 'Emma']) should return a list containing Ava, Emma, Olivia, and Sophia in any order.

Implement the function find_roots to find the roots of the quadratic equation: ax² + bx + c = 0. The function should return a tuple containing roots in any order. If the equation has only one solution, the function should return that solution as both elements of the tuple. 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.

Binary search tree (BST) is a binary tree where the value of each node is larger or equal to the values in all the nodes in that node's left subtree and is smaller than the values in all the nodes in that node's right subtree.

Write a function that, efficiently with respect to time used, checks if a given binary search tree contains a given value.

For example, for the following tree:

• n1 (Value: 1, Left: null, Right: null)
• n2 (Value: 2, Left: n1, Right: n3)
• n3 (Value: 3, Left: null, Right: null)

Call to contains(n2, 3) should return True since a tree with root at n2 contains number 3.

A playlist is considered a repeating playlist if any of the songs contain a reference to a previous song in the playlist. Otherwise, the playlist will end with the last song which points to None.

Implement a function is_repeating_playlist that, efficiently with respect to time used, returns true if a playlist is repeating or false if it is not.

For example, the following code prints "True" as both songs point to each other.

first = Song("Hello")
second = Song("Eye of the tiger")
    
first.next_song(second)
second.next_song(first)
    
print(first.is_repeating_playlist())

Write a function that, when passed a list and a target sum, returns, efficiently with respect to time used, two distinct zero-based indices of any two of the numbers, whose sum is equal to the target sum. If there are no two numbers, the function should return None.

For example, find_two_sum([3, 1, 5, 7, 5, 9], 10) should return a single tuple containing any of the following pairs of indices:

• 0 and 3 (or 3 and 0) as 3 + 7 = 10
• 1 and 5 (or 5 and 1) as 1 + 9 = 10
• 2 and 4 (or 4 and 2) as 5 + 5 = 10

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.

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(lambda x: x * 3, lambda x: x + 1, lambda x: x / 2) then calling the returned function with 3 should return 5.0.

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.

The LeagueTable class tracks the score of each player in a league. After each game, the player records their score with the record_result function. 

The player's rank in the league is calculated using the following logic:

1. The player with the highest score is ranked first (rank 1). The player with the lowest score is ranked last.
2. If two players are tied on score, then the player who has played the fewest games is ranked higher.
3. If two players are tied on score and number of games played, then the player who was first in the list of players is ranked higher.

Implement the player_rank function that returns the player at the given rank.

For example:

table = LeagueTable(['Mike', 'Chris', 'Arnold'])
table.record_result('Mike', 2)
table.record_result('Mike', 3)
table.record_result('Arnold', 5)
table.record_result('Chris', 5)
print(table.player_rank(1))

All players have the same score. However, Arnold and Chris have played fewer games than Mike, and as Chris is before Arnold in the list of players, he is ranked first. Therefore, the code above should display "Chris".

Implement function count_numbers that accepts a sorted list of unique integers and, efficiently with respect to time used, counts the number of list elements that are less than the parameter less_than.

For example, count_numbers([1, 3, 5, 7], 4) should return 2 because there are two list elements less than 4.

A TrainComposition is built by attaching and detaching wagons from the left and the right sides, efficiently with respect to time used.

For example, if we start by attaching wagon 7 from the left followed by attaching wagon 13, again from the left, we get a composition of two wagons (13 and 7 from left to right). Now the first wagon that can be detached from the right is 7 and the first that can be detached from the left is 13.

Implement a TrainComposition that models this problem.

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.

As a part of the route planner, the route_exists method is used as a quick filter if the destination is reachable, before using more computationally intensive procedures for finding the optimal route.

The roads on the map are rasterized and produce a matrix of boolean values - True if the road is present or False if it is not. The roads in the matrix are connected only if the road is immediately left, right, below or above it.

Finish the route_exists method so that it returns True if the destination is reachable or False if it is not. The from_row and from_column parameters are the starting row and column in the map_matrix. The to_row and to_column are the destination row and column in the map_matrix. The map_matrix parameter is the above mentioned matrix produced from the map.

For example, for the given rasterized map, the code below should return True since the destination is reachable:

map_matrix = [
    [True, False, False],
    [True, True, False],
    [False, True, True]
];

route_exists(0, 0, 2, 2, map_matrix)