human
What is a the Python code you requested given the following:
Solve in Python:
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.
assistant
#Here is the Python code you requested.

class Solution:
    def racecar(self, target: int) -> int:
        # from answers:
        # Dojkstra's
        '''
        K = target.bit_length() + 1
        barrier = 1 << K
        pq = [(0, target)]
        dist = [float('inf')] * (2 * barrier + 1)
        dist[target] = 0

        while pq:
            steps, targ = heapq.heappop(pq)
            if dist[targ] > steps: continue

            for k in range(K+1):
                walk = (1 << k) - 1
                steps2, targ2 = steps + k + 1, walk - targ
                if walk == targ: steps2 -= 1 #No \"R\" command if already exact

                if abs(targ2) <= barrier and steps2 < dist[targ2]:
                    heapq.heappush(pq, (steps2, targ2))
                    dist[targ2] = steps2

        return dist[0]
        '''
        # Dynamic programming
        dp = [0, 1, 4] + [float('inf')] * target
        for t in range(3, target + 1):
            k = t.bit_length()
            if t == 2**k - 1:
                dp[t] = k
                continue
            for j in range(k - 1):
                dp[t] = min(dp[t], dp[t - 2**(k - 1) + 2**j] + k - 1 + j + 2)
            if 2**k - 1 - t < t:
                dp[t] = min(dp[t], dp[2**k - 1 - t] + k + 1)
        return dp[target]