LeetCode Problem Workspace

Race Car

Race Car is a dynamic programming problem where the goal is to find the shortest sequence of instructions to reach a target position.

category

1

Topics

code_blocks

4

Code langs

hub

3

Related

Practice Focus

Hard · State transition dynamic programming

bolt

Answer-first summary

Race Car is a dynamic programming problem where the goal is to find the shortest sequence of instructions to reach a target position.

Interview AiBox logo

Ace coding interviews with Interview AiBox guidance for State transition dynamic programming

Try AiBox Copilotarrow_forward

To solve the Race Car problem, you'll need to use dynamic programming to find the shortest sequence of instructions that leads to the target position. The car's position and speed are dynamically adjusted by the commands 'A' for acceleration and 'R' for reverse. Understanding the state transitions and leveraging optimal substructure properties is key to an efficient solution.

Problem Statement

You are driving a car that starts at position 0 with speed 1 on an infinite number line. The car can move in either direction, and its movement is controlled by a series of instructions. The instructions are 'A' for acceleration and 'R' for reverse. The car’s position and speed evolve as follows: after 'A', the car accelerates and its speed increases by 1. After 'R', the car reverses, and its speed becomes negative. The car can end up in negative positions, so understanding state transitions is crucial.

Given a target position, you need to determine the shortest sequence of instructions that gets the car from position 0 to the target. Your task is to implement a function that returns the minimal instruction sequence. The sequence should use 'A' for acceleration and 'R' for reverse, and the goal is to minimize the number of instructions required to reach the target.

Examples

Example 1

Input: target = 3

Output: 2

The shortest instruction sequence is "AA". Your position goes from 0 --> 1 --> 3.

Example 2

Input: target = 6

Output: 5

The shortest instruction sequence is "AAARA". Your position goes from 0 --> 1 --> 3 --> 7 --> 7 --> 6.

Constraints

  • 1 <= target <= 104

Solution Approach

State Transition Dynamic Programming

The core of the solution involves using dynamic programming to track the car's position and speed at each step. Each state is represented by the car's position and speed, and the transitions between states depend on the acceleration ('A') and reverse ('R') commands. The solution requires exploring possible state transitions and identifying the optimal sequence of commands to reach the target position.

Breadth-First Search (BFS) Approach

A BFS approach can be used to explore all possible states systematically. Starting from the initial state (position 0, speed 1), the BFS iterates over all possible actions ('A' and 'R') and explores the resulting positions and speeds. Each new state is added to the queue and processed, ensuring the shortest path is found by expanding states in increasing order of steps.

Memoization and Optimization

To improve efficiency, memoization can be applied to store already visited states. This prevents redundant calculations and speeds up the process. Additionally, limiting the state space by considering only the relevant positions and speeds within a reasonable range helps optimize the solution and avoid unnecessary exploration.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time and space complexity depend on the approach used to explore states. In the BFS approach, the time complexity is proportional to the number of possible states, which can be large but is limited by the target and speed. Memoization helps reduce redundant calculations, improving space efficiency. The space complexity primarily depends on the size of the state space, which can be large but manageable with optimizations.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1

#### Python3

1
2
3
4
5
6
7
8
9
10
11
12
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]
Race Car Solution: State transition dynamic programming | LeetCode #818 Hard