Car Fleet II

Hard

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ArraysStacks

There are n cars traveling at different speeds in the same direction along a one-lane road. You are given an array cars of length n, where cars[i] = [position_i, speed_i] represents:

position_i is the distance between the i^th car and the beginning of the road in meters. It is guaranteed that position_i < position_i+1.
speed_i is the initial speed of the i^th car in meters per second.

For simplicity, cars can be considered as points moving along the number line. Two cars collide when they occupy the same position. Once a car collides with another car, they unite and form a single car fleet. The cars in the formed fleet will have the same position and the same speed, which is the initial speed of the slowest car in the fleet.

Return an array answer, where answer[i] is the time, in seconds, at which the i^th car collides with the next car, or -1 if the car does not collide with the next car. Answers within 10^-5 of the actual answers are accepted.

Example 1:

Input: cars = [[1,2],[2,1],[4,3],[7,2]]
Output: [1.00000,-1.00000,3.00000,-1.00000]
Explanation: After exactly one second, the first car will collide with the second car, and form a car fleet with speed 1 m/s. After exactly 3 seconds, the third car will collide with the fourth car, and form a car fleet with speed 2 m/s.

Example 2:

Input: cars = [[3,4],[5,4],[6,3],[9,1]]
Output: [2.00000,1.00000,1.50000,-1.00000]

Constraints:

1 <= cars.length <= 10⁵
1 <= position_i, speed_i <= 10⁶
position_i < position_i+1

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum number of cars that can be in the input array, and what are the possible ranges for the position and speed of each car?
Can a car have a speed of zero, or a negative position value?
If a car never collides with any car ahead of it, what should be the return value for that car's collision time?
Are the initial positions of the cars guaranteed to be unique and sorted in ascending order, or do I need to handle unsorted input and potential position duplicates?
If multiple collisions occur for a single car, should I return the time of the very first collision only?

Brute Force Solution

Approach

The brute force approach for the car fleet problem involves simulating each car's movement and checking for collisions. For each car, we consider all possible future collisions with cars ahead of it. We then determine the time of the earliest collision, if any, and use that to calculate the impact force.

Here's how the algorithm would work step-by-step:

Start with the first car.
Consider all the cars in front of the first car one by one.
For each car ahead, calculate the time it would take for the first car to collide with it.
If the cars won't collide (because the car in front is faster), move on to the next car ahead.
If they will collide, remember the collision time and the car it collides with.
Repeat this process for all cars in front to find the earliest collision time for the first car.
If a collision occurs, calculate the impact force based on the speeds and masses of the colliding cars.
If no collision occurs, there is no impact force for that car.
Repeat this whole process for the second car, then the third car, and so on, until you've considered all the cars.
The resulting impact forces are the answer.

Code Implementation

def car_fleet_ii_brute_force(cars):
    number_of_cars = len(cars)
    impact_forces = [0.0] * number_of_cars

    for current_car_index in range(number_of_cars):
        earliest_collision_time = float('inf')
        colliding_car_index = -1

        # Iterate through all cars ahead
        for next_car_index in range(current_car_index + 1, number_of_cars):
            current_car_position = cars[current_car_index][0]
            current_car_speed = cars[current_car_index][1]
            next_car_position = cars[next_car_index][0]
            next_car_speed = cars[next_car_index][1]

            # If the car ahead is faster, no collision will occur
            if current_car_speed <= next_car_speed:
                continue

            time_to_collide = (next_car_position - current_car_position) / (current_car_speed - next_car_speed)

            # Determine the earliest collision
            if time_to_collide < earliest_collision_time:
                earliest_collision_time = time_to_collide
                colliding_car_index = next_car_index

        # Calculate impact force if a collision occurs
        if colliding_car_index != -1:
            current_car_mass = cars[current_car_index][2]
            next_car_mass = cars[colliding_car_index][2]

            relative_speed = cars[colliding_car_index][1] - cars[current_car_index][1]

            # Ensuring no division by zero if total mass is 0
            if current_car_mass + next_car_mass == 0:
              impact_forces[current_car_index] = 0.0
            else:
              impact_forces[current_car_index] = (current_car_mass * relative_speed) / (current_car_mass + next_car_mass)

    return impact_forces

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n cars. For each car, it compares it with every car ahead of it to find the earliest collision time. In the worst-case scenario, the algorithm performs a collision check for each pair of cars. This results in roughly n * (n-1) / 2 comparisons, which is approximately n squared divided by two. Therefore, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach outlined involves iterating through the cars and calculating collision times and forces directly. It does not mention the creation of any auxiliary data structures like lists, hash maps, or stacks to store intermediate results beyond the input array. Only a few variables might be needed to track the earliest collision time and the corresponding car during the nested loops. Therefore, the auxiliary space used is constant, independent of the number of cars (N).

Optimal Solution

Approach

The goal is to figure out when each car fleet will collide and what the cost of that collision will be. We use a stack to keep track of potentially colliding cars and determine collision times efficiently.

Here's how the algorithm would work step-by-step:

Go through the cars from right to left (from the last car to the first car).
For each car, we want to find the next car in front of it that it will collide with, if any.
Use a stack to store the cars that are in front of the current car and are potentially going to be collided with.
Before adding the current car to the stack, check if it will collide with the cars already in the stack. To do this, compare the time it would take for the current car to collide with the car at the top of the stack with the time it would take for the car at the top of the stack to collide with the car behind it (also in the stack).
If the current car will collide with the car at the top of the stack sooner than that car collides with the car behind it, it means that the car at the top of the stack will be hit by the current car before it can hit the car in front of it. In that case, remove the car at the top of the stack because it is no longer relevant.
Repeat this process of checking and removing cars from the stack until you find a car in the stack that the current car will not collide with first, or the stack is empty.
If the stack is not empty, it means the current car will collide with the car at the top of the stack. Calculate the time it takes for the collision to occur, and the cost of the collision (the cost of the car at the top of the stack). If the stack is empty, the current car will not collide with any other car.
Store the collision time and cost for the current car.
Add the current car to the stack. This car is now a potential collision target for cars behind it.
After processing all the cars, you'll have the collision time and cost for each car.

Code Implementation

def car_fleet_ii(cars):
    number_of_cars = len(cars)
    collision_times = [-1.0] * number_of_cars
    stack = []

    for i in range(number_of_cars - 1, -1, -1):
        while stack:
            # Collision time between current car and car on stack top
            position_i, speed_i, cost_i = cars[i]
            position_j, speed_j, cost_j = cars[stack[-1]]

            if speed_i <= speed_j:
                stack.pop()
                continue

            time_to_collide_ij = (position_j - position_i) / (speed_i - speed_j)

            if len(stack) >= 2:
                # Consider the next car on the stack.
                position_k, speed_k, cost_k = cars[stack[-2]]
                position_j, speed_j, cost_j = cars[stack[-1]]
                if speed_j <= speed_k:
                    time_to_collide_jk = float('inf')
                else:
                    time_to_collide_jk = (position_k - position_j) / (speed_j - speed_k)

                # We check if car j would collide with k before i collides with j.
                if time_to_collide_ij >= time_to_collide_jk:
                    # Car j will collide with car k first, so we can remove it.
                    stack.pop()
                    continue
            
            # Collision found and stack is not empty.
            collision_times[i] = time_to_collide_ij
            break

        stack.append(i)

    return collision_times

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n cars once, from right to left. Inside the loop, a stack is used to potentially remove elements. Each car is added to the stack at most once and removed at most once. Thus, while there's a while loop inside the main loop, the while loop's operations are bounded by the number of elements, so it does not contribute n^2. Thus, the overall time complexity is dominated by the single pass through the array of cars, resulting in O(n) time complexity.

Space Complexity

O(N) – The primary auxiliary space usage comes from the stack which, in the worst case, could contain all the cars. This occurs when no collisions happen, and each car is added to the stack. Therefore, the stack's size is directly proportional to the number of cars, N. Since we store at most N car indices in the stack, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty 'cars' array	Return an empty result array as there are no cars to form fleets.
Single car in 'cars' array	Return an array containing -1 since a single car cannot form a fleet.
All cars have the same initial position	The car with the smallest speed will be the only one to arrive, and other cars behind will form fleets with it; calculate the collision time and append -1 for others.
All cars have zero speed	Cars will never collide, so return -1 for each car.
Very large car positions or speeds, causing potential integer overflow when calculating arrival times	Use long data types for storing arrival times to prevent overflow.
Cars are extremely close to each other initially, and the speeds cause near-instant collisions resulting in floating-point precision issues	Use a reasonable epsilon value to compare floating-point numbers and avoid precision issues.
Cars are sorted in reverse order of initial positions	Ensure the stack processes the cars in the correct order, popping elements until the current car collides sooner than the top of the stack or the stack is empty.
Maximum number of cars allowed (array size constraint), leading to potential performance issues (stack size)	The algorithm's time complexity is O(n) (due to single pass) where n is cars length and space complexity is also O(n) due to stack; This is generally acceptable for reasonable constraints, but consider optimizing with more advanced data structures if performance becomes a major bottleneck with extreme constraints.

Empty 'cars' array

How to Handle:

Return an empty result array as there are no cars to form fleets.

Single car in 'cars' array

How to Handle:

Return an array containing -1 since a single car cannot form a fleet.

All cars have the same initial position

How to Handle:

The car with the smallest speed will be the only one to arrive, and other cars behind will form fleets with it; calculate the collision time and append -1 for others.

All cars have zero speed

How to Handle:

Cars will never collide, so return -1 for each car.

Very large car positions or speeds, causing potential integer overflow when calculating arrival times

How to Handle:

Use long data types for storing arrival times to prevent overflow.

Cars are extremely close to each other initially, and the speeds cause near-instant collisions resulting in floating-point precision issues

How to Handle:

Use a reasonable epsilon value to compare floating-point numbers and avoid precision issues.

Cars are sorted in reverse order of initial positions

How to Handle:

Ensure the stack processes the cars in the correct order, popping elements until the current car collides sooner than the top of the stack or the stack is empty.

Maximum number of cars allowed (array size constraint), leading to potential performance issues (stack size)

How to Handle:

The algorithm's time complexity is O(n) (due to single pass) where n is cars length and space complexity is also O(n) due to stack; This is generally acceptable for reasonable constraints, but consider optimizing with more advanced data structures if performance becomes a major bottleneck with extreme constraints.