<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:
        que = [(0, 1, 0)] #num_ops, screen_num, buffer_num
        seen = defaultdict()
        seen[(1, 0)] = 0
        while que:
            num_ops, cur, buffer = heappop(que)
            if cur == n:
                return num_ops
            else:
                #two ops
                if cur <= n and ((cur, cur) not in seen or seen[(cur, cur)] > num_ops + 1):
                    seen[(cur, cur)] = num_ops + 1
                    heappush(que, (num_ops + 1, cur, cur))
                if cur + buffer <= n and ((cur + buffer, buffer) not in seen or seen[(cur + buffer, buffer)] > num_ops + 1):
                    #paste operation, growing screen string only
                    seen[(cur + buffer, buffer)] = num_ops + 1
                    heappush(que, (num_ops + 1,  cur + buffer, buffer))
            #print(que)
        return n - 1

    def minSteps(self, n: int) -> int:
        if n == 1: return 0
        if n == 2: return 2
        min_op = n
        for i in range(1, (n//2)+1):
            if n%i != 0: continue
            min_op = min(min_op, self.minSteps(i)+(n//i))
        return min_op


    def minSteps(self, n:int) -> int:
        @cache
        def dp(cur, copy):
            if cur == n: return 0
            ans = inf
            if copy and cur + copy <= n: ans = min(ans, 1 + dp(cur+copy, copy))
            if copy < cur and 2 * cur <= n: ans = min(ans, 1 + dp(cur, cur))
            return ans

        return dp(1, 0)

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