<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) {
        // dp[i][j]: i+1 'A's on screen, j 'A's on clipboard
        auto dp = vector<vector<int> >();
        for (int i = 0; i < n; i++) {
            dp.emplace_back(i + 2, 100000);
        }
        dp[0][0] = 0;
        dp[0][1] = 1;

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i + 1 - j; j++) {
                // j 'A's on clipboard i+1-j 'A' already on screen
                dp[i][j] = dp[i - j][j] + 1;
            }
            dp[i][i + 1] = *min_element(dp[i].begin(), dp[i].end()) + 1;
        }

        return *min_element(dp[n - 1].begin(), dp[n - 1].end());
    }
};
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