<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>

from collections import deque
class Solution:
    def minSteps(self, n: int) -> int:
        kMax = 2_000_000_000
        dp = []
        for i in range(n + 1):
            d = []
            for j in range(n + 1):
                d.append(kMax)
            dp.append(d)
        dp[1][0] = 0
        dp[1][1] = 1
        for i in range(1, n+1):
            for j in range(n + 1):
                if i + j < n + 1:
                    dp[i + j][j] = min(dp[i+j][j], dp[i][j] + 1)
                dp[i][i] = min(dp[i][i], dp[i][j] + 1)
        ans = kMax
        for i in range(n + 1):
            if dp[n][i] < ans:
                ans = dp[n][i]
        return ans
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