# Solving Leetcode Interviews in Seconds with AI: Race Car


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "818" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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):  When you get an instruction 'A', your car does the following:  	 position += speed speed *= 2   When you get an instruction 'R', your car does the following: 	 If your speed is positive then speed = -1 otherwise speed = 1  	Your position stays the same.  For example, after commands "AAR", your car goes to positions 0 --> 1 --> 3 --> 3, and your speed goes to 1 --> 2 --> 4 --> -1. Given a target position target, return the length of the shortest sequence of instructions to get there.   Example 1:  Input: target = 3 Output: 2 Explanation:  The shortest instruction sequence is "AA". Your position goes from 0 --> 1 --> 3.  Example 2:  Input: target = 6 Output: 5 Explanation:  The shortest instruction sequence is "AAARA". Your position goes from 0 --> 1 --> 3 --> 7 --> 7 --> 6.    Constraints:  1 <= target <= 104  

	# Explanation
	*   **BFS Approach:** Use Breadth-First Search (BFS) to explore possible instruction sequences. The state in BFS is defined by (position, speed, steps).
*   **State Space Pruning:** Optimize BFS by pruning states that are too far from the target or have already been visited with fewer steps. Specifically, limit positions and speeds within a reasonable range to avoid infinite loops and unnecessary exploration.
*   **Bidirectional Search (Optimization):** In principle, a bidirectional search from the target can also work, but it does not add much optimization in this case. Therefore, a standard BFS approach is preferred for clarity.

*   **Runtime Complexity:** O(target * log(target)), **Storage Complexity:** O(target * log(target))

	
	# Code
	```python
	from collections import deque

def shortest_instructions(target):
    """
    Finds the length of the shortest sequence of instructions to reach the target.

    Args:
        target: The target position.

    Returns:
        The length of the shortest sequence of instructions.
    """

    q = deque([(0, 1, 0)])  # (position, speed, steps)
    visited = set()
    visited.add((0, 1))

    while q:
        pos, speed, steps = q.popleft()

        if pos == target:
            return steps

        # Option 1: Accelerate
        new_pos = pos + speed
        new_speed = speed * 2
        if abs(new_pos) < 2 * target and (new_pos, new_speed) not in visited:
            q.append((new_pos, new_speed, steps + 1))
            visited.add((new_pos, new_speed))

        # Option 2: Reverse
        new_speed = -1 if speed > 0 else 1
        if abs(pos) < 2 * target and (pos, new_speed) not in visited:
            q.append((pos, new_speed, steps + 1))
            visited.add((pos, new_speed))

    return -1  # Should not happen given the problem constraints
	```