<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 f(int num, int lastcp, int tar, vector<vector<int>> &dp) {
    if (num == tar) return 0;
    if (num + lastcp > tar) return 2000;
    
    if (dp[num][lastcp] != -1) return dp[num][lastcp];
    int cp = 2 + f(num + num, num, tar, dp);
    int p = 1 + f(num + lastcp, lastcp, tar, dp);

    return dp[num][lastcp] = min(cp, p);
    }

    int minSteps(int n) {
        if (n == 1) return 0;
        vector<vector<int>> dp(n + 1, vector<int>(n, -1));
        return f(2, 1, n, dp) + 2;
    }

};
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