<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:
        '''
        2 moves
            - copy all
            - paste characters which are copied last time
        
        'A'
        n= 3 
        

        first: number of A
        mode:
            0: current number is not copy
            1: copied

        '1,1'
        '2,1'
     '2,2'  '3,1'
    
        base case: return 1 
        logic: 
            if curr_a  == curr_register: dp[a][r] = 1 + dp[a+1][r]
            else:
                dp[a][r] = min(dp[a+1][r], dp[a][a])
        '''

        cache = {}
        if n == 1:
            return 0

        def dp(curr_a, curr_reg):
            if curr_a == n:
                return 0
            elif curr_a >= n:
                return 1000
            elif (curr_a, curr_reg) in cache:
                return cache[(curr_a, curr_reg)]
            
            if curr_a == curr_reg:
                cache[(curr_a, curr_reg)] = 1 + dp(curr_a*2, curr_reg)
            else:
                print(curr_a, curr_reg)
                cache[(curr_a, curr_reg)] = 1 + min(dp(curr_a+curr_reg, curr_reg), dp(curr_a, curr_a))
            
            return cache[(curr_a, curr_reg)]
        
        return dp(1,1)+1
              
33653