<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 solve(int t,int f,int cc,int n,vector<vector<int>>&dp)
        {
           if(t==n)
               return 0;
       
          if(t>n)
              return 1000;
        
       //  if(dp[t][f]!=-1)
          //  return dp[t][f];
        
        int val=0;
        if(f==0)
            {
               val=1+solve(t,1,t,n,dp);
            }
          else
val=min(1+solve(t+cc,0,0,n,dp),1+solve(t+cc,1,cc,n,dp));
        
       // cout<<t<<" "<<val<<" ";
        return dp[t][f]=val;
        }
    int minSteps(int n) {
        vector<vector<int>>dp(n+1,vector<int>(2,-1));
        return solve(1,0,0,n,dp);
    }
};
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