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Race Car

Hard

Race Car

Hard

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:

Is the target distance a positive integer?
Are there any constraints on the range of possible positions we can reach or the number of instructions we can execute?
If it is impossible to reach the target distance, what should I return?
Can the car overshoot the target, or does it need to land exactly on it?
Is the target distance small enough to fit within the integer range (e.g., 32 or 64 bits)?

Brute Force Solution

Approach

The brute force strategy for the race car problem involves exploring every single possible sequence of instructions to reach the target position. It's like trying every combination of 'A' (accelerate) and 'R' (reverse) commands until we stumble upon the shortest one that gets us there. We keep track of all the possibilities and then pick the best one.

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

Start with an empty instruction sequence and the race car at the starting position and speed.
Try adding either an 'A' (accelerate) or an 'R' (reverse) command to the sequence.
After each command, update the car's position and speed based on the command.
If the car's position is equal to the target, record the current instruction sequence as a potential solution.
If the car's position has gone past the target position in either direction, or the instruction sequence has become too long, discard that sequence.
Repeat the process of adding 'A' or 'R' to all possible instruction sequences until all paths have either reached the target, gone past the target, or exceeded a maximum sequence length.
From all the recorded instruction sequences that reach the target, choose the shortest sequence. That sequence represents the solution.

Code Implementation

def race_car_brute_force(target_position):
    shortest_path = None

    queue = [((0, 1, ""), 0)]

    while queue:
        (current_position, current_speed, current_instructions), instruction_count = queue.pop(0)

        # If we overshoot the target, prune the branch
        if abs(current_position) > abs(target_position) * 2 or instruction_count > 20:
            continue

        # Found a path that is shorter than the current shortest path
        if current_position == target_position:
            if shortest_path is None or len(current_instructions) < len(shortest_path):
                shortest_path = current_instructions
            continue

        # Explore accelerating
        new_position = current_position + current_speed
        new_speed = current_speed * 2
        queue.append(((new_position, new_speed, current_instructions + "A"), instruction_count + 1))

        #Reverse direction. If the current speed is positive, make it negative. If negative, positive.
        new_speed = -1 if current_speed > 0 else 1
        queue.append(((current_position, new_speed, current_instructions + "R"), instruction_count + 1))

    return shortest_path

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible sequence of 'A' and 'R' instructions up to a certain length. In the worst-case scenario, we might need to explore sequences of length proportional to the target position 'n'. Since each position in the sequence can have two possible values ('A' or 'R'), the total number of sequences we explore grows exponentially with the target 'n'. Therefore, the total number of operations is proportional to 2 raised to the power of n, which leads to a time complexity of O(2^n).

Space Complexity

O(2^N) – The brute force approach explores all possible instruction sequences of 'A' and 'R'. In the worst case, to find the shortest path, we might explore a large portion of the possible sequences up to a certain length N, where N is related to how far the car may stray from the target due to repeated acceleration or reversals before finding a solution (or determining there is none). The algorithm implicitly uses a queue or stack-like structure to keep track of these sequences to explore, resulting in a space complexity proportional to the number of possible sequences explored. Thus, the space complexity grows exponentially with the maximum sequence length, approximating O(2^N), as each step has two choices.

Optimal Solution

Approach

The best way to solve this problem is to use a method called Breadth-First Search. We explore possible moves in layers, always looking for the shortest path to the target position.

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

Think of each position as a place on a graph, and each instruction ('A' for accelerate, 'R' for reverse) as a way to move to a new place.
Start at position zero with speed one.
Explore all the places you can reach in one move (either accelerate or reverse).
Then, explore all the places you can reach in two moves from the starting point.
Keep going, layer by layer, until you reach the target position.
Crucially, as you explore, remember the number of steps needed to get to each position so far.
If you find a shorter way to reach a position you've already visited, update the number of steps needed.
Once you reach the target position, the number of steps you recorded is the shortest path.

Code Implementation

from collections import deque

def racecar(target):

    queue = deque([(0, 1, 0)])
    visited = set()
    visited.add((0, 1))

    while queue:
        position, speed, steps = queue.popleft()

        if position == target:
            return steps

        # Move with accelerate instruction
        next_position = position + speed
        next_speed = speed * 2
        if (next_position, next_speed) not in visited and \
           0 < next_position < 2 * target:

            queue.append((next_position, next_speed, steps + 1))
            visited.add((next_position, next_speed))

        # Move with reverse instruction
        next_speed = -1 if speed > 0 else 1

        # Reversing direction might lead to optimal path.
        if (position, next_speed) not in visited and 0 < position < 2 * target:

            queue.append((position, next_speed, steps + 1))
            visited.add((position, next_speed))

    return -1 # Should never happen

Big(O) Analysis

Time Complexity

O(n log n) – In Breadth-First Search (BFS), we explore reachable states (position, speed) layer by layer until we find the target. In the worst case, we might visit all possible states before reaching the target. The number of reachable positions grows linearly with the number of instructions 'n'. For each position, the maximum speed can also be related to 'n'. We use a queue to process each position, and queue operations take O(1) on average. Furthermore, we use a visited set, and looking up/inserting to it typically takes O(log n) time using efficient data structures. Therefore, the overall time complexity is bounded by O(n log n), considering the number of states and cost of operations.

Space Complexity

O(N^2) – The breadth-first search algorithm uses a queue to store positions and speeds to explore. The maximum position reachable in the search space can be approximated by the target value N. Since the speed can be at most N as well, the queue may contain position-speed pairs up to O(N * N), which simplifies to O(N^2). Additionally, a visited set or similar structure may be used to avoid revisiting states, potentially storing O(N^2) elements as well. Therefore, the auxiliary space complexity is O(N^2).

Edge Cases

Case	How to Handle
Target distance is zero.	Return 0 immediately since no moves are needed.
Target distance is negative.	Assume the problem specifies non-negative distance, or handle by mirroring the solution for the corresponding positive distance.
Integer overflow when calculating the next position or speed.	Use a data type with a larger range (e.g., long) or check for overflow before the operation.
Very large target distance (close to the maximum integer value).	Consider the memory and time complexity when exploring the search space, potentially causing a timeout or memory error; use iterative approach and limit the search depth.
Initial speed is zero.	Start with an initial speed of 1 or handle it as a specific initial condition depending on the problem description.
The car can reach the target distance and immediately stop.	The algorithm should prioritize reaching the target as quickly as possible, even if it involves some overshooting and reversals.
Pathological case where the optimal path involves many reversals before reaching the target.	Employ a heuristic or pruning strategy within the search algorithm (e.g., BFS or Dijkstra's) to avoid exploring unnecessarily long or complex paths; limit maximum queue size in BFS.
Floating-point representation errors accumulating in position or speed.	Use integer arithmetic instead of floating-point numbers to represent position and speed, or set an acceptable tolerance for error; this is usually not applicable, as problem parameters will use integers

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:

Is the target distance a positive integer?
Are there any constraints on the range of possible positions we can reach or the number of instructions we can execute?
If it is impossible to reach the target distance, what should I return?
Can the car overshoot the target, or does it need to land exactly on it?
Is the target distance small enough to fit within the integer range (e.g., 32 or 64 bits)?

Brute Force Solution

Approach

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

Start with an empty instruction sequence and the race car at the starting position and speed.
Try adding either an 'A' (accelerate) or an 'R' (reverse) command to the sequence.
After each command, update the car's position and speed based on the command.
If the car's position is equal to the target, record the current instruction sequence as a potential solution.
If the car's position has gone past the target position in either direction, or the instruction sequence has become too long, discard that sequence.
Repeat the process of adding 'A' or 'R' to all possible instruction sequences until all paths have either reached the target, gone past the target, or exceeded a maximum sequence length.
From all the recorded instruction sequences that reach the target, choose the shortest sequence. That sequence represents the solution.

Code Implementation

def race_car_brute_force(target_position):
    shortest_path = None

    queue = [((0, 1, ""), 0)]

    while queue:
        (current_position, current_speed, current_instructions), instruction_count = queue.pop(0)

        # If we overshoot the target, prune the branch
        if abs(current_position) > abs(target_position) * 2 or instruction_count > 20:
            continue

        # Found a path that is shorter than the current shortest path
        if current_position == target_position:
            if shortest_path is None or len(current_instructions) < len(shortest_path):
                shortest_path = current_instructions
            continue

        # Explore accelerating
        new_position = current_position + current_speed
        new_speed = current_speed * 2
        queue.append(((new_position, new_speed, current_instructions + "A"), instruction_count + 1))

        #Reverse direction. If the current speed is positive, make it negative. If negative, positive.
        new_speed = -1 if current_speed > 0 else 1
        queue.append(((current_position, new_speed, current_instructions + "R"), instruction_count + 1))

    return shortest_path

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

The best way to solve this problem is to use a method called Breadth-First Search. We explore possible moves in layers, always looking for the shortest path to the target position.

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

Think of each position as a place on a graph, and each instruction ('A' for accelerate, 'R' for reverse) as a way to move to a new place.
Start at position zero with speed one.
Explore all the places you can reach in one move (either accelerate or reverse).
Then, explore all the places you can reach in two moves from the starting point.
Keep going, layer by layer, until you reach the target position.
Crucially, as you explore, remember the number of steps needed to get to each position so far.
If you find a shorter way to reach a position you've already visited, update the number of steps needed.
Once you reach the target position, the number of steps you recorded is the shortest path.

Code Implementation

from collections import deque

def racecar(target):

    queue = deque([(0, 1, 0)])
    visited = set()
    visited.add((0, 1))

    while queue:
        position, speed, steps = queue.popleft()

        if position == target:
            return steps

        # Move with accelerate instruction
        next_position = position + speed
        next_speed = speed * 2
        if (next_position, next_speed) not in visited and \
           0 < next_position < 2 * target:

            queue.append((next_position, next_speed, steps + 1))
            visited.add((next_position, next_speed))

        # Move with reverse instruction
        next_speed = -1 if speed > 0 else 1

        # Reversing direction might lead to optimal path.
        if (position, next_speed) not in visited and 0 < position < 2 * target:

            queue.append((position, next_speed, steps + 1))
            visited.add((position, next_speed))

    return -1 # Should never happen

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Target distance is zero.	Return 0 immediately since no moves are needed.
Target distance is negative.	Assume the problem specifies non-negative distance, or handle by mirroring the solution for the corresponding positive distance.
Integer overflow when calculating the next position or speed.	Use a data type with a larger range (e.g., long) or check for overflow before the operation.
Very large target distance (close to the maximum integer value).	Consider the memory and time complexity when exploring the search space, potentially causing a timeout or memory error; use iterative approach and limit the search depth.
Initial speed is zero.	Start with an initial speed of 1 or handle it as a specific initial condition depending on the problem description.
The car can reach the target distance and immediately stop.	The algorithm should prioritize reaching the target as quickly as possible, even if it involves some overshooting and reversals.
Pathological case where the optimal path involves many reversals before reaching the target.	Employ a heuristic or pruning strategy within the search algorithm (e.g., BFS or Dijkstra's) to avoid exploring unnecessarily long or complex paths; limit maximum queue size in BFS.
Floating-point representation errors accumulating in position or speed.	Use integer arithmetic instead of floating-point numbers to represent position and speed, or set an acceptable tolerance for error; this is usually not applicable, as problem parameters will use integers