<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:
// vector<vector<int>>dp;
unordered_map<long long,unordered_map<long long,unordered_map<long long,long long>>>dp;
    long long f(int num,int c,int isc, int tar){
long long copy=INT_MAX;
long long paste=INT_MAX;
// cout<<num<<" "<<c<<"\n";
if(c>tar)return INT_MAX;
if(num>tar)return INT_MAX;

if(num==tar)return 0;
if(dp.count(num)&&dp[num].count(c)&&dp[num][c].count(isc))return dp[num][c][isc];
if(!isc){

        copy=1+(long long)f(num,num,1,tar);
}
if(num+c<=tar&&c>0){

         paste=1+(long long)f(num+c,c,0,tar);
}
cout<<copy<<" "<<paste<<"\n";

return dp[num][c][isc]=copy<=paste?copy:paste;
    }
    int minSteps(int n) {
        // dp=vector<vector<int>>(n,vector<int>(n+1))
return f(1,0,0,n);
    }
};
105603