What is the main pattern used to solve the Race Car problem?

The main pattern used is state transition dynamic programming, where each state represents the car's position and speed, and transitions depend on the 'A' and 'R' commands.

How does BFS help in solving the Race Car problem?

BFS helps by systematically exploring all possible states from the starting position, ensuring the shortest path to the target is found.

What are some optimization techniques for the Race Car problem?

Memoization is commonly used to store already visited states, reducing redundant calculations and speeding up the solution. Limiting the state space also helps optimize the approach.

Can the Race Car problem be solved without dynamic programming?

While dynamic programming is an optimal approach, BFS can also be used to explore states and find the shortest sequence of instructions.

What is the expected time complexity of solving the Race Car problem?

The time complexity depends on the BFS approach and memoization optimizations, with complexity tied to the number of states explored.

#818

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

赛车

你的赛车可以从位置 0 开始，并且速度为 +1 ，在一条无限长的数轴上行驶。赛车也可以向负方向行驶。赛车可以按照由加速指令 'A' 和倒车指令 'R' 组成的指令序列自动行驶。当收到指令 'A' 时，赛车这样行驶： position += speed speed *= 2 当收到指令 'R' 时，…

动态规划

题目描述

你的赛车可以从位置 0 开始，并且速度为 +1 ，在一条无限长的数轴上行驶。赛车也可以向负方向行驶。赛车可以按照由加速指令 'A' 和倒车指令 'R' 组成的指令序列自动行驶。

当收到指令 'A' 时，赛车这样行驶：
- position += speed
- speed *= 2
当收到指令 'R' 时，赛车这样行驶：
- 如果速度为正数，那么speed = -1
- 否则 speed = 1
当前所处位置不变。

例如，在执行指令 "AAR" 后，赛车位置变化为 0 --> 1 --> 3 --> 3 ，速度变化为 1 --> 2 --> 4 --> -1 。

给你一个目标位置 target ，返回能到达目标位置的最短指令序列的长度。

示例 1：

输入：target = 3
输出：2
解释：
最短指令序列是 "AA" 。
位置变化 0 --> 1 --> 3 。

示例 2：

输入：target = 6
输出：5
解释：
最短指令序列是 "AAARA" 。
位置变化 0 --> 1 --> 3 --> 7 --> 7 --> 6 。

提示：

1 <= target <= 10⁴

lightbulb

解题思路

方法一：动态规划

设 $dp[i]$ 表示到达位置 $i$ 的最短指令序列的长度。答案为 $dp[target]$ 。

对于任意位置 $i$ ，都有 $2^{k-1} \leq i \lt 2^k$ ，并且我们可以有三种方式到达位置 $i$ ：

如果 $i$ 等于 $2^k-1$ ，那么我们可以直接执行 $k$ 个 A 指令到达位置 $i$ ，此时 $dp[i] = k$ ；
否则，我们可以先执行 $k$ 个 A 指令到达位置 $2^k-1$ ，然后执行 R 指令，剩余距离为 $2^k-1-i$ ，此时 $dp[i] = dp[2^k-1-i] + k + 1$ ；我们也可以先执行 $k-1$ 个 A 指令到达位置 $2^{k-1}-1$ ，然后执行 R 指令，接着执行 $j$ （其中 $0 \le j \lt k$ ）个 A，再执行 R，剩余距离为 $i - 2^{k-1} + 2^j$ ，此时 $dp[i] = dp[i - 2^{k-1} + 2^j] + k - 1 + j + 2$ 。求出 $dp[i]$ 的最小值即可。

时间复杂度 $O(n \log n)$ ，其中 $n$ 为 $target$ 。

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class Solution:
    def racecar(self, target: int) -> int:
        dp = [0] * (target + 1)
        for i in range(1, target + 1):
            k = i.bit_length()
            if i == 2**k - 1:
                dp[i] = k
                continue
            dp[i] = dp[2**k - 1 - i] + k + 1
            for j in range(k - 1):
                dp[i] = min(dp[i], dp[i - (2 ** (k - 1) - 2**j)] + k - 1 + j + 2)
        return dp[target]

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Can the candidate identify and optimize the state transitions for minimal steps?
question_mark
Does the candidate consider the use of dynamic programming or BFS to solve the problem efficiently?
question_mark
How well does the candidate handle space and time optimization, such as with memoization?

warning

常见陷阱

外企场景

error
Not considering negative speeds or positions when calculating the car’s movements.
error
Failing to optimize the search space with memoization or state space reduction.
error
Overcomplicating the solution by not using dynamic programming or BFS to explore the shortest path efficiently.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the target is very large? Consider the impact of expanding the state space.
arrow_right_alt
What if the car could change its speed more than once per step? This would introduce a different dynamic in the state transition.
arrow_right_alt
What if the target position was negative? The solution would need to account for reversing and going in the negative direction.

help

常见问题

外企场景

继续练习

#813 最大平均值和的分组 #823 带因子的二叉树 #808 分汤