<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:
    # next = prev + copy
    # n [1, 10^5]
    # 6 -> [1, 2, 4, 6] 2+2+1=5   6 -> [1,2,3,6] 2+1+2=5
    # dp[i][j] get to i through increasing j, prev number is i-j
    # if i-j == j, we use copy all (2 operations)
    # if i-j != j, we use paste only (1 operations)
    # i-j >= 1, so dp[i][i] == 0
    # dp[2][1] = dp[1][1] + 2 = 2
    # dp[3][1] = dp[2][1] + 1 = 3, dp[3][2] = dp[1][2] = 0 
    def minSteps(self, n: int) -> int:
        # edge case, one char is already on screen
        if n == 1:
            return 0
        dp = [[float('inf')]*(n+1) for i in range(n+1)]
        dp[1][1] = 0
        # [f,f,f,f]
        # [f,0,f,f]
        # [f,2,f,f]
        # [f,3,f,f]
        # [f,4,4,f]
        # [f,5,f,f]
        # [f,6,5,5]
        for i in range(2,n+1):
            # dp[i][j], prev number is i-j>=j, i>=2j
            for j in range(1, i//2+1):
                prev = i - j
                if prev == j:
                    dp[i][j] = min(dp[prev]) + 2
                else:
                    dp[i][j] = dp[prev][j] + 1
        return min(dp[n])

        
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