<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:
    def minSteps(self, n: int) -> int:
        @lru_cache(maxsize=1000)
        def dfs(cur_len_of_A, cur_len_of_copy):
            if cur_len_of_A == n:
                return 0
            
            if cur_len_of_A > 1000: # highest constraint
                return 1000
            
            # one option is to copy cur A's and then paste them, 
            # which takes 2 ops and cur copy len will be cur_len_of_A in clipboard
            # and cur len of A's will be doubled
            copy_then_paste = 2 + dfs(cur_len_of_A * 2, cur_len_of_A)

            # other option is to paste cur_copy without no more copy => only 1 op
            # so cur len of A's will be increased by cur copy len and 
            # copy len will stay the same as no futher copying takes place
            paste_only = 1 + dfs(cur_len_of_A + cur_len_of_copy, cur_len_of_copy)

            return min(copy_then_paste, paste_only)
        
        return dfs(1, 1) + 1 if n>1 else 0 # plus 1 for intial copy of A's 
21098