<p>There is only one character <code>&#39;A&#39;</code> on the screen of a notepad. You can perform one of two operations on this notepad for each step:</p>

<ul>
	<li>Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).</li>
	<li>Paste: You can paste the characters which are copied last time.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the minimum number of operations to get the character</em> <code>&#39;A&#39;</code> <em>exactly</em> <code>n</code> <em>times on the screen</em>.</p>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, we have one character &#39;A&#39;.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get &#39;AA&#39;.
In step 3, we use Paste operation to get &#39;AAA&#39;.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:

    int n;
    int minSteps(int n) {

        if(n == 1) return 0;
        this->n = n;
        unordered_map<string,int> map;
        return 1 + minStepsHelper(1,1,map);
    }

    int minStepsHelper(int currLen, int pasteLen, unordered_map<string,int>& map){
        string key = to_string(currLen) + "," + to_string(pasteLen);
        if(map.find(key) != map.end()) return map[key];
        if(currLen == n) return 0;

        if(currLen > 1000) return 1000;

        // copy and paste
        int option1 = 2 + minStepsHelper(currLen * 2, currLen,map);
        // paste
        int option2 = 1 + minStepsHelper(currLen + pasteLen, pasteLen,map);

        map[key] = min(option1,option2);
        return map[key];
    }
};
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