Need to practice your Python programming skills for an upcoming job interview? Try solving these Python interview questions that test knowledge of lists, iteration, recursion, sorting, and other skills. We’ll provide feedback on your answers, and you can use a hint if you get stuck.

These Python interview questions are examples of real tasks used by employers to screen job candidates such as Python developers, back-end developers, full-stack developers, and others that require knowledge of the Python programming language and its native data structures and language constructs.

**1. File Owners**

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']}*.

- Example case: Wrong answer
- Each owner has a single file: Wrong answer
- Various files: Wrong answer
- Function group_by_owners is called more than one time: Wrong answer

**2. Quadratic Equation**

Implement the function *find_roots* to find the roots of the quadratic equation: ax^{2} + 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 2x^{2} + 10x + 8 = 0 are -1 and -4.

**3. Binary Search Tree**

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.

- Example case: Wrong answer
- Correctness: Wrong answer
- Performance test on a large tree: Wrong answer

**4. Song**

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())
```

- Example case: Wrong answer
- If playlist repeats, last song is followed by first song: Wrong answer
- If playlist repeats, starting songs can be skipped: Wrong answer
- Performance test on a large playlist: Wrong answer

**5. Two Sum**

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

- Example case: Wrong answer
- Distinct numbers with and without solutions: Wrong answer
- Duplicate numbers with and without solutions: Wrong answer
- Performance test with a large list of numbers: Wrong answer

**6. League Table**

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:

- The player with the highest score is ranked first (rank 1). The player with the lowest score is ranked last.
- If two players are tied on score, then the player who has played the fewest games is ranked higher.
- 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".

- Example case: Wrong answer
- Players have different scores: Wrong answer
- Players tied by score: Wrong answer
- Players tied by games played: Wrong answer

**7. Sorted Search**

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.

- Example case: Wrong answer
- Various small lists: Wrong answer
- Performance test when sorted_list contains less_than: Wrong answer
- Performance test when sorted_list doesn't contain less_than: Wrong answer

**8. Train Composition**

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.

- Example Case: Wrong answer
- Several Wagons: Wrong answer
- Performance test with a large number of wagons: Wrong answer

**9. Route Planner**

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)
```

- Example case: Wrong answer
- Various routes from top-left to bottom-right corner: Wrong answer
- Various routes, starts, and destinations: Wrong answer
- Performance test on large map: Wrong answer

**10. Ice Cream Machine**

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']].

- Example case: Wrong answer
- Various ingredients and one topping: Wrong answer
- Various ingredients and toppings: Wrong answer
- No ingredients and no toppings: Wrong answer

**11. Merge Names**

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.

- Example case: Wrong answer
- Each array has distinct names: Wrong answer
- Each array has duplicate names: Wrong answer
- Arrays have some names in common: Wrong answer

**12. Pipeline**

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.

- Example case: Wrong answer
- Various functions: Wrong answer
- Various data types: Wrong answer