<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:
    unordered_map<string, int> um;
    int solve(int n, int copy, int total_len, int copy_len){
        if(total_len > n){
            return 1e6;
        }
        if(n == total_len)
            return 0;
        string key = to_string(copy) + " " + to_string(total_len) + " " + to_string(copy_len);
        if(um.find(key) != um.end()){
            return um[key];
        }
        int result;
        if(copy){
            result = 1 + solve(n, 0, total_len, total_len);
        }else{
            int case_1 = 1 + solve(n, 0, total_len + copy_len, copy_len);
            int case_2 = 1 + solve(n, 1, total_len + copy_len, total_len + copy_len);            
            result = min(case_1, case_2);
        }
        return um[key] = result;
    }
    int minSteps(int n) {
        if(n == 1)
            return 0;
        return solve(n, 1, 1, 1);
    }
};
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