> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/graph-search/818.race-car.md).

# 818.Race-Car

## 818. Race Car

## 题目地址

<https://leetcode.com/problems/race-car/>

## 题目描述

```
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.

Now for some target position, say 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.

Note:
1 <= target <= 10000.
```

## 代码

### Approach #1 BFS Solution

```java
public int racecar(int target) {
  Deque<int[]> queue = new LinkedList<>();
  queue.offerLast(new int[] {0, 1}); // starts from position 0 with speed 1
  Set<String> visited = new HashSet<>();
  visited.add(0 + " " + 1);

  for (int level = 0; !queue.isEmpty(); level++) {
    for (int k = queue.size(); k > 0; k--) {
      int[] cur = queue.pollFirst(); // cur[0] is position; cur[1] is speed
      if (cur[0] == target) {
        return level;
      }

      int[] nxt = new int[] {cur[0] + cur[1], cur[1] << 1};  // accelerate instruction
      String key = (nxt[0] + " " + nxt[1]);

      if (!visited.contains(key) && 0 < nxt[0] && nxt[0] < (target << 1)) {
        queue.offerLast(nxt);
        visited.add(key);
      }
      nxt = new int[] {cur[0], cur[1] > 0 ? -1 : 1}; // reverse instruction
      key = (nxt[0] + " " + nxt[1]);

      if (!visited.contains(key) && 0 < nxt[0] && nxt[0] < (target << 1)) {
        queue.offerLast(nxt);
        visited.add(key);
      }
    }
  }
  return -1;
}
```

### #2 BFS

```java
class Solution {
  class CarInfo{
      int pos, speed;
      public CarInfo(int p, int s) {
          pos = p;
          speed = s;
      }
  }

  public int racecar(int target) {
    Set<String> visited = new HashSet<>();
    String begin = 0 + "/" + 1;
    visited.add(begin);
    Queue<CarInfo> queue = new LinkedList<>();
    queue.add(new CarInfo(0,1));
    int level = 0;
    while (!queue.isEmpty()) {
      int size = queue.size();
      for(int i = 0; i < size; i++) {
          CarInfo cur = queue.poll();
          if (cur.pos == target) return level;
          String s1 = (cur.pos + cur.speed) + "/" + (cur.speed * 2);
          String s2 = cur.pos + "/" + (cur.speed > 0 ? -1 : 1);
          if (Math.abs(cur.pos + cur.speed - target) < target && !visited.contains(s1)) {
              visited.add(s1);
              queue.add(new CarInfo(cur.pos + cur.speed, cur.speed * 2));
          }
          if (Math.abs(cur.pos - target) < target && !visited.contains(s2)) {
              visited.add(s2);
              queue.add(new CarInfo(cur.pos, cur.speed > 0 ? -1 : 1));
          }
      }

      level++;
    }
    return -1;
  }

}
```

### Approach #2 Dynamic Programming

`Consider two general cases for number i with bit_length n.` `i==2^n-1, this case, n is the best way`2^(n-1)-1\<i<2^n-1, there are two ways to arrive at i: `Use n A to arrive at 2^n-1 first, then use R to change the direction, by here there are n+1 operations (n A and one R), then the remaining is same as dp[2^n-1-i]`Use n-1 A to arrive at 2^(n-1)-1 first, then R to change the direction, use m A to go backward, then use R to change the direction again to move forward, by here there are n-1+2+m=n+m+1 operations (n-1 A, two R, m A) , current position is 2^(n-1)-1-(2^m-1)=2^(n-1)-2^m, the remaining operations is same as dp\[i-(2^(n-1)-1)+(2^m-1)]=dp\[i-2^(n-1)+2^m)].\`

```java
int[] dp = new int[10001];
  public int racecar(int t) {
    if (dp[t] > 0) return dp[t];
    int n = (int)(Math.log(t) / Math.log(2)) + 1;
    if (1 << n == t + 1) {
        dp[t] = n;
    } else {
        dp[t] = racecar((1 << n) - 1 - t) + n + 1;
        for (int m = 0; m < n - 1; ++m) {
            dp[t] = Math.min(dp[t], racecar(t - (1 << (n - 1)) + (1 << m)) + n + m + 1);
        }
    }
    return dp[t];
  }
```

### Top Down DP

```java
public int racecar(int target) {
    int[] dp = new int[target + 1];
    Arrays.fill(dp, 1, dp.length, -1);
    return racecar(target, dp);
}

private int racecar(int target, int[] dp) {
  if (dp[target] >= 0) {
      return dp[target];
  }

  dp[target] = Integer.MAX_VALUE;

  int m = 1, j = 1;

  for (; j < target; j = (1 << ++m) - 1) {
    for (int q = 0, p = 0; p < j; p = (1 << ++q) - 1) {
      dp[target] = Math.min(dp[target],  m + 1 + q + 1 + racecar(target - (j - p), dp));
    }
  }

  dp[target] = Math.min(dp[target], m + (target == j ? 0 : 1 + racecar(j - target, dp)));

  return dp[target];
}
```

### Bottom-up DP

```java
public int racecar(int target) {
    int[] dp = new int[target + 1];

    for (int i = 1; i <= target; i++) {
        dp[i] = Integer.MAX_VALUE;

        int m = 1, j = 1;

        for (; j < i; j = (1 << ++m) - 1) {
            for (int q = 0, p = 0; p < j; p = (1 << ++q) - 1) {
                dp[i] = Math.min(dp[i], m + 1 + q + 1 + dp[i - (j - p)]);
            }
        }

        dp[i] = Math.min(dp[i], m + (i == j ? 0 : 1 + dp[j - i]));
    }

    return dp[target];
}
```


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter, and the optional `goal` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/graph-search/818.race-car.md?ask=<question>&goal=<endgoal>
```

`ask` is the immediate question: it should be specific, self-contained, and written in natural language.
`goal` is optional and describes the broader end goal you are ultimately trying to accomplish on behalf of the user. GitBook uses it to tailor the answer towards what is most useful for that goal.

The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.