<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 minSteps(int n) {

        if(n==1) return 0;
        multiset<pair<int,pair<int,int>>> ms;
        ms.insert({2,{2,1}});
        while(!ms.empty()){
            int oper = ms.begin()->first;
            int len = ms.begin()->second.first;
            int prev_len = ms.begin()->second.second;
            ms.erase(ms.begin());

            if(len == n) return oper;

            if(len+prev_len <= n)
                ms.insert({oper+1,{len+prev_len,prev_len}});
            if(2*len <= n)
                ms.insert({oper+2,{2*len,len}});
        }
        return -1;
        
    }
};
/*

(number of operations,length,last copy length)
(2,2,1)
f(o,l,cl) -> f(o+1,l+cl,cl)
          -> f(o+2,2*l,l)
*/
98098