How do you solve Race Car on LeetCode?

Race Car (LeetCode #818) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the impact of acceleration and reversal on position and speed is crucial.

What is the brute force solution for Race Car?

The brute force approach for Race Car is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries all possible sequences of instructions to reach the target position. It explores every combination of 'A' and 'R' until it finds the shortest path to the target. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Race Car?

Race Car uses the following concepts: Dynamic Programming. The recommended patterns to study are: Breadth-First Search (BFS), Dynamic Programming.

What are the interview tips for Race Car?

Always clarify the problem and constraints before diving into coding. Think about edge cases, such as when the target is very close or far. Explain your thought process clearly as you code; it shows your understanding.

What are common mistakes when solving Race Car?

Not considering the impact of reversing on future moves. Failing to track visited states, leading to infinite loops.

#818

Race Car

Hard

Dynamic ProgrammingBreadth-First Search (BFS)Dynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to minimize the number of instructions. It builds a table that tracks the minimum steps required to reach each position based on previous calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a queue to perform BFS and a set to track visited states.
2Step 2: Start from position 0 with speed 1 and initialize the instruction count.
3Step 3: For each state, calculate the next positions based on 'A' and 'R', and add them to the queue if they haven't been visited.

solution.py15 lines

1from collections import deque
2
3def racecar(target):
4    queue = deque([(0, 1, 0)])
5    visited = set()
6    while queue:
7        pos, speed, steps = queue.popleft()
8        if pos == target:
9            return steps
10        if (pos, speed) in visited:
11            continue
12        visited.add((pos, speed))
13        queue.append((pos + speed, speed * 2, steps + 1))
14        queue.append((pos, -1 if speed > 0 else 1, steps + 1))
15    return -1

ℹ

Complexity note: The time complexity is O(n) because we explore each position and speed combination only once. The space complexity is O(n) due to the queue and visited set storing states.

1Understanding the impact of acceleration and reversal on position and speed is crucial.
2The problem can be visualized as navigating a state space where each state is defined by position and speed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.