Race Car
<p>Your car starts at position <code>0</code> and speed <code>+1</code> on an infinite number line. Your car can go into negative positions. Your car drives automatically according to a sequence of instructions <code>&#39;A&#39;</code> (accelerate) and <code>&#39;R&#39;</code> (reverse):</p>

<ul>
	<li>When you get an instruction <code>&#39;A&#39;</code>, your car does the following:

    <ul>
    	<li><code>position += speed</code></li>
    	<li><code>speed *= 2</code></li>
    </ul>
    </li>
    <li>When you get an instruction <code>&#39;R&#39;</code>, your car does the following:
    <ul>
    	<li>If your speed is positive then <code>speed = -1</code></li>
    	<li>otherwise <code>speed = 1</code></li>
    </ul>
    Your position stays the same.</li>

</ul>

<p>For example, after commands <code>&quot;AAR&quot;</code>, your car goes to positions <code>0 --&gt; 1 --&gt; 3 --&gt; 3</code>, and your speed goes to <code>1 --&gt; 2 --&gt; 4 --&gt; -1</code>.</p>

<p>Given a target position <code>target</code>, return <em>the length of the shortest sequence of instructions to get there</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> target = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
The shortest instruction sequence is &quot;AA&quot;.
Your position goes from 0 --&gt; 1 --&gt; 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> target = 6
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
The shortest instruction sequence is &quot;AAARA&quot;.
Your position goes from 0 --&gt; 1 --&gt; 3 --&gt; 7 --&gt; 7 --&gt; 6.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= target &lt;= 10<sup>4</sup></code></li>
</ul>
Dynamic Programming
class Solution {
public:
    int racecar(int target) {
        vector<int> dp(target + 1);
        for (int i = 1; i <= target; ++i) {
            int k = 32 - __builtin_clz(i);
            if (i == (1 << k) - 1) {
                dp[i] = k;
                continue;
            }
            dp[i] = dp[(1 << k) - 1 - i] + k + 1;
            for (int j = 0; j < k; ++j) {
                dp[i] = min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
            }
        }
        return dp[target];
    }
};