<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 {
public:
    int minSteps(int n) {

        if(n == 1)
        {
            return 0;
        }

        int maxCopied = n / 2 + 1;

        vector<vector<int>> dp(n);
        for(int i = 0; i < n; ++i)
        {
            dp[i].resize(min(i + 1, maxCopied), 2000);
        }

        for(int i = 0; i < maxCopied; ++i)
        {
            dp[n-1][i] = 0;
        }

        for(int printed = n - 1; printed >= 0; --printed)
        {
            for(int copied = 0; copied < dp[printed].size() - 1; ++copied)
            {
                dp[printed][copied] = min(dp[printed][copied], dp[printed].back() + 1);
            }

            for(int copied = 0; copied < dp[printed].size(); ++copied)
            {
                if(printed - copied - 1 < 0 || dp[printed - copied - 1].size() <= copied)
                {
                    continue;
                }

                dp[printed - copied - 1][copied] = min(dp[printed - copied - 1][copied], 1 + dp[printed][copied]);
            }
        }

        return 1 + dp[0][0];
    }
};
68078