<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>

class Solution {
    vector<int> dp;
public:
    int racecar(int target) {
        dp = vector<int>(300001, -1);
        return racecarInternal(target);
    }

    int racecarInternal(int target) {
        if(dp[target] != -1) {
            return dp[target];
        }
        if(target == 0 || target == 1) {
            return target;
        }
        
        /* reach nearest 2^(n-1) - 1 value before target and put on 2 reverses 
        (so that speed is changed to 1) and then solve for target-(2^(n-1)-1)
        or after 1 reverse, do (n-2) accelerate operations and then calculate
        or
        reach 2^n - 1 value after target, put one reverse and solve for (2^n-1-target)
        */

        int pos = 0;
        int speed = 1;
        int steps = 0;
        while(pos < target) {
            // accelerate
            pos += speed;
            speed *= 2;
            steps++;
        }

        if(pos == target) {
            return steps;
        }

        // pos > target

        int before = pos/2;
        // take one reverse, and then reach target
        int stepsFromAfter = steps+1+racecarInternal(pos-target);

        // take 2 reverses and then try reaching target from before
        int stepsFromBefore = steps+1+racecarInternal(target-before);

        dp[target] = min(stepsFromAfter,stepsFromBefore);
        speed = 1;
        
        for(int i=0;i<steps-2;i++) {
            before = before-speed;
            speed = speed*2;
            dp[target] = min(dp[target], steps+i+2+racecarInternal(target-before));
        }
        return dp[target];
    }
};
79898