<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:
//  cpy all
//  paste all copied
int dp[1001][1001][3];
int f(int cpy,int left,int pre,int n){
    if(left==0)return 0;
    if(left<0)return 1e9;
    if(dp[cpy][left][pre]!=-1)return dp[cpy][left][pre];
    if(cpy==0)
        return (1+f(1,left,1,n));
    int ans=1e9;
    if(pre)
    ans=min(ans,1+f(cpy,left-cpy,0,n));
    if(!pre){
    ans=min(ans,1+f(cpy,left-cpy,0,n));
    ans=min(ans,1+f(n-left,left,1,n));
    }
    return dp[cpy][left][pre]=ans;
}
    int minSteps(int n) {
        // if we do copy then we have to do paste else do not cpy
        memset(dp,-1,sizeof dp);
        return f(0,n-1,0,n);
    }
};
21171