<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 {
    int recur(int curr, int paste, int &n, unordered_map<string, int>& memo) {
        if(curr == n){
            return 0;
        }

        if(curr > n){
            return 1000;
        }
        string key = to_string(curr) + "," + to_string(paste);
        if(memo.find(key) != memo.end()){
            return memo[key];
        }

        int val1 = 2 + recur(curr * 2, curr, n, memo);
        int val2 = 1 + recur(curr + paste, paste, n, memo);

        memo[key] =  min(val1, val2);
        return memo[key];
    }
public:
    int minSteps(int n) {
        if(n == 1){
            return 0;
        }
        
        unordered_map<string, int> memo;
        return 1 + recur(1, 1, n, memo);
    }
};
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