<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:
    # [1, 2, 4, 8, 16, 32]
    def minSteps(self, n: int) -> int:
        if n == 1:
            return 0
        self.memo = [[0 for _ in range(n)] for _ in range(n)]
        return 1 + self.helper(1, 1, n)


    def helper(self, cur_val, clipboard, n):
        if cur_val == n:
            return 0
        
        if cur_val > n:
            return n

        if self.memo[cur_val][clipboard]:
            return self.memo[cur_val][clipboard]

        op_1 = self.helper(cur_val + clipboard, clipboard, n) + 1
        op_2 = self.helper(cur_val * 2, cur_val, n) + 2


        self.memo[cur_val][clipboard] = min(op_1, op_2)
        return self.memo[cur_val][clipboard]
        
        

        

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