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Race Car

Difficulty: Hard

Problem Description

Your car starts at position 0 and speed +1 on an infinite number line. Your car can go into negative positions. Your car drives automatically according to a sequence of instructions 'A' (accelerate) and 'R' (reverse). Given a target position, return the length of the shortest sequence of instructions to get there.

Key Insights

  • The car can accelerate (A) to double its speed and move forward, or reverse (R) to change its speed direction without changing position.
  • The problem can be visualized as a state space where each state is defined by the current position and speed of the car.
  • A breadth-first search (BFS) approach can be applied to explore all possible sequences of instructions, keeping track of visited states to avoid cycles.
  • The target can be reached optimally by considering both accelerating and reversing at each step.

Space and Time Complexity

Time Complexity: O(N) where N is the maximum distance needed to reach the target. This is due to the BFS exploring all possible positions. Space Complexity: O(N) for storing the states in the queue and the visited set.

Solution

The solution employs the breadth-first search (BFS) algorithm to explore all possible positions and speeds of the car. Each state is represented by the current position and speed. The BFS expands nodes by applying the 'A' and 'R' commands and enqueues new states until the target position is reached. A set is used to track visited states to prevent redundant exploration.

Code Solutions

from collections import deque

def racecar(target):
    queue = deque([(0, 1, 0)])  # (position, speed, steps)
    visited = set((0, 1))
    
    while queue:
        position, speed, steps = queue.popleft()
        
        # Check if we reached the target
        if position == target:
            return steps
        
        # Option 1: Accelerate
        new_position = position + speed
        new_speed = speed * 2
        if (new_position, new_speed) not in visited and abs(new_position) <= 2 * target:
            visited.add((new_position, new_speed))
            queue.append((new_position, new_speed, steps + 1))
        
        # Option 2: Reverse
        new_speed = -1 if speed > 0 else 1
        if (position, new_speed) not in visited:
            visited.add((position, new_speed))
            queue.append((position, new_speed, steps + 1))
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