<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>

#include<bits/stdc++.h>
class Solution {
public:
    int dp[1001][1001][3];
    
    int solve(int i, int prev, int op, int n){
        if(i==n) return 0;
        if(i>n) return 1e9;
        if(dp[i][prev][op]!=-1) return dp[i][prev][op];
        int ans = INT_MAX;
        int copy, paste;
        copy = paste = INT_MAX;
        if(prev==0) ans = 1 + solve(i, i, 1, n);
        if(prev>0){
            if(op==1){
                paste = 1 + solve(i+prev, prev, 2, n);
            }
            else if(op==2){
                copy = 1 + solve(i, i, 1, n);
                paste = 1 + solve(i+prev, prev, 2, n);
            }
            ans = min(copy, paste);
        }
        
        return dp[i][prev][op]=ans;
    }
    int minSteps(int n) {
        memset(dp, -1, sizeof(dp));
        return solve(1, 0, 0, n);
    }
};

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