<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 {
     unordered_map<long int,unordered_map<long int,unordered_map<long int,long int> > >dp;
public:
    long int minStepsUtil(int n,int prev,int curr)
    {
        if(dp.find(n)!=dp.end()&&dp[n].find(prev)!=dp[n].end()&&dp[n][prev].find(curr)!=dp[n][prev].end())
        return dp[n][prev][curr];
        
        if(n==0)
        return dp[n][prev][curr]=0;
        if(n<0)
        return dp[n][prev][curr]=INT_MAX;
        if(prev == -1)
        return dp[n][prev][curr]=2+minStepsUtil(n-1,1,2);
        else
        return dp[n][prev][curr]=min(2+minStepsUtil(n-curr,curr,2*curr),1+minStepsUtil(n-prev,prev,curr+prev));
    }
    long int minSteps(int n) {
       
         return minStepsUtil(n-1,-1,1);

        
    }
};




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