<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:
        cache = {}
        def dfs(charCount, clipboardCount, opCount):
            # print(charCount, clipboardCount, opCount)
            if charCount > n:
                return float('inf')

            if charCount == n:
                print(cache)
                return opCount

            if (charCount, clipboardCount) in cache:
                return cache[(charCount, clipboardCount)] + opCount

            minOpCount = float('inf')

            # 1. Copy only - charCount != clipboardCount
            if charCount != clipboardCount:
                # print("Copy")
                minOpCount = min(minOpCount, dfs(charCount, charCount, opCount + 1))

            #2. Paste
            # print("Paste")
            if clipboardCount != 0:
                minOpCount = min(minOpCount, dfs(charCount + clipboardCount, clipboardCount, opCount + 1))

            if minOpCount != float("inf"):
                cache[(charCount, clipboardCount)] = minOpCount - opCount
                print(charCount, minOpCount - opCount)

            return minOpCount

        res = dfs(1, 0, 0)
        return res

            
        
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