<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:
        if n == 1:
            return 0

        # dp[i][j]:
        # how many operations needed to obtain
        # i 'A's on the board
        # j 'A's in the register
        dp = [[math.inf] * (k+1) for k in range(n+1)]
        dp[1][0] = 0
        
        for n_on_screen in range(1, n+1):
            for n_in_register in range(0, n_on_screen+1):
                # copy to register
                dp[n_on_screen][n_on_screen] = min(
                    dp[n_on_screen][n_on_screen], 
                    dp[n_on_screen][n_in_register]+1,
                )
                # paste to board
                if n_on_screen + n_in_register <= n:
                    dp[n_on_screen + n_in_register][n_in_register] = min(
                        dp[n_on_screen + n_in_register][n_in_register],
                        dp[n_on_screen][n_in_register] + 1,
                    )

        return min(dp[-1])
        
20168