Placing Racing Car

Interview Question

Imagine you are developing a racing game. You need to implement a system that determines the placement of each car in the race at any given moment. The race track is a circular track of a fixed length. Each car has a unique ID and a current distance traveled from the starting line. Design a function that takes a list of cars, where each car is represented by its ID and distance traveled, and returns the current placement of each car, considering the circular nature of the track. For example, if the track length is 1000 meters and a car has traveled 1200 meters, its actual position on the track is 200 meters. Consider edge cases such as multiple cars having the same position. Can you implement a function that efficiently calculates and assigns placements to the cars based on their positions on the track?

Naive Solution

A straightforward approach is to first calculate the actual position of each car on the track by taking the distance traveled modulo the track length. Then, sort the cars based on their positions to determine their placement. This approach is easy to understand but might not be the most efficient.

def calculate_placements_naive(cars, track_length):
    """Calculates the placement of each car in the race (naive approach).

    Args:
        cars (list of tuples): List of cars, where each car is represented by (car_id, distance_traveled).
        track_length (int): The length of the race track.

    Returns:
        dict: A dictionary where the key is car_id and the value is the car's placement.
    """
    positions = {}
    for car_id, distance_traveled in cars:
        positions[car_id] = distance_traveled % track_length

    sorted_cars = sorted(positions.items(), key=lambda item: item[1], reverse=True)

    placements = {}
    rank = 1
    for car_id, _ in sorted_cars:
        placements[car_id] = rank
        rank += 1

    return placements

Optimal Solution

To improve efficiency, we can use a similar approach but optimize the sorting process. The key improvement is maintaining the relative order in case of ties, and using python's built in sort function with the correct key. Here's how we can implement it in Python:

def calculate_placements_optimal(cars, track_length):
    """Calculates the placement of each car in the race (optimal approach).

    Args:
        cars (list of tuples): List of cars, where each car is represented by (car_id, distance_traveled).
        track_length (int): The length of the race track.

    Returns:
        dict: A dictionary where the key is car_id and the value is the car's placement.
    """
    positions = []
    for car_id, distance_traveled in cars:
        positions.append((car_id, distance_traveled % track_length))

    # Sort cars based on their positions
    sorted_cars = sorted(positions, key=lambda x: x[1], reverse=True)

    placements = {}
    rank = 1
    if sorted_cars:
      placements[sorted_cars[0][0]] = rank
      for i in range(1, len(sorted_cars)):
        if sorted_cars[i][1] != sorted_cars[i-1][1]:
          rank = i + 1
        placements[sorted_cars[i][0]] = rank

    return placements

Big(O) Run-time Analysis

Naive Solution:
- Calculating positions: O(n), where n is the number of cars. We iterate through each car in the cars list once to calculate the modulo and store in the dictionary positions. This is a simple linear operation. We do this before the sort.
- Sorting: O(n log n), where n is the number of cars. Python's sorted() function uses Timsort, which has an average and worst-case time complexity of O(n log n). This sorts the cars based on their calculated positions. This is more significant than O(n) for larger n. So, it dominates the time complexity.
- Assigning placements: O(n), where n is the number of cars. Once the cars are sorted, we iterate through them again to assign placements. This is a linear operation, but less significant than the sort.
- Overall: O(n log n) due to the sorting step.
Optimal Solution:
- Calculating positions: O(n), where n is the number of cars. Just like in the naive solution, we iterate through the cars list to calculate the positions, resulting in O(n)..
- Sorting: O(n log n), where n is the number of cars. This is still the most expensive operation, with Python's sorted() using Timsort.
- Assigning placements: O(n), where n is the number of cars. Iterating to assign rank takes O(n).
- Overall: O(n log n), dominated by the sorting algorithm.

Big(O) Space Usage Analysis

Naive Solution:
- positions: O(n), where n is the number of cars. Stores the positions of each car.
- sorted_cars: O(n), where n is the number of cars. Stores the sorted list of cars.
- placements: O(n), where n is the number of cars. Stores the final placements of each car.
- Overall: O(n)
Optimal Solution:
- positions: O(n), where n is the number of cars. This array holds the car IDs and calculated track positions.
- sorted_cars: O(n), where n is the number of cars. Stores sorted car data.
- placements: O(n), where n is the number of cars. This dict holds the final car placements.
- Overall: O(n)

Edge Cases

Empty list of cars:
- Return an empty dictionary.
Track length of zero:
- Handle this case by returning 1 for all cars since they are all at the same spot.
Negative distance traveled:
- The modulo operator will handle negative distances correctly in Python. The position will be distance_traveled % track_length and python handles negative modulos correctly.
Cars with the same position:
- The solution correctly handles cars with the same position by assigning them the same rank.
Large number of cars:
- The algorithm's time complexity of O(n log n) will handle a large number of cars efficiently.
Cars traveling extremely large distances:
- The modulo operator ensures that the position remains within the track length, even for very large distances.

Here's the code, modified to handle edge cases such as empty list of cars or a track length of 0.

def calculate_placements_edge_cases(cars, track_length):
    """Calculates the placement of each car in the race, handling edge cases.

    Args:
        cars (list of tuples): List of cars, where each car is represented by (car_id, distance_traveled).
        track_length (int): The length of the race track.

    Returns:
        dict: A dictionary where the key is car_id and the value is the car's placement.
    """
    if not cars:
        return {}

    if track_length == 0:
        placements = {}
        for car_id, _ in cars:
            placements[car_id] = 1
        return placements

    positions = []
    for car_id, distance_traveled in cars:
        positions.append((car_id, distance_traveled % track_length))

    sorted_cars = sorted(positions, key=lambda x: x[1], reverse=True)

    placements = {}
    rank = 1
    if sorted_cars:
      placements[sorted_cars[0][0]] = rank
      for i in range(1, len(sorted_cars)):
        if sorted_cars[i][1] != sorted_cars[i-1][1]:
          rank = i + 1
        placements[sorted_cars[i][0]] = rank

    return placements

How would you implement a placement system for cars in a racing game with a circular track?