<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 dp[2048][2048];

    int solve(int targetLen, int notepadLen, int currentLen) {
        if(currentLen == targetLen) {
            return dp[notepadLen][currentLen] = 0;
        }

        if(currentLen > targetLen) {
            return dp[notepadLen][currentLen] = INT_MAX/2;
        }

        if(dp[notepadLen][currentLen] != -1) {
            return dp[notepadLen][currentLen];
        }

        int copy_paste = solve(targetLen, currentLen, currentLen*2) + 2;

        // int paste = INT_MAX;
        // if(notepadLen > 0) {
        int paste = solve(targetLen, notepadLen, notepadLen + currentLen) + 1;
        // }

        return dp[notepadLen][currentLen] =  min(copy_paste, paste);
    }

    int minSteps(int n) {
        if(n == 1) return 0;

        memset(dp, -1, sizeof(dp));

        return 1 + solve(n, 1, 1);
    }
};
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