<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) {
        vector<int> dp(n + 1, 0);

        dp[1] = 0;
        if (n > 1) dp[2] = 2;

        for (int i = 3; i <= n; ++i) {
            vector<int> arrOfDiv;  
                
            for (int j = 1; j <= i / 2; ++j) {
                if (i % j == 0) {
                    arrOfDiv.push_back(j);
                }
            }
            
            int mini = INT_MAX;
            for (int j = 0; j < arrOfDiv.size(); ++j) {
                mini = min(dp[arrOfDiv[j]] + (i / arrOfDiv[j]), mini);
            }
            
            dp[i] = mini;
        }

        for (auto i : dp) cout << i << " ";
        
        return dp[n];
    }
};

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