<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) {
        if (n == 1) return 0;
        else {
            map<pair<int, int>, int> cache = {};
            return 2 + complete(2, n, 1, cache);
        }
    }

    int complete(int cur, int n, int clipboard, map<pair<int, int>, int>& cache) {
        if (cur == n) return 0;
        else if (cur > n) return 10000;
        else if (cache.find({cur, clipboard}) != cache.end()) return cache[{cur, clipboard}];
        else {
            int n1 = 10000, n2 = 10000;
            if (cur + clipboard <= n) {
                n1 = complete(cur+clipboard, n, clipboard, cache) + 1;
            }
            if (cur * 2 <= n) {
                n2 = 2 + complete(cur*2, n, cur, cache);
            }
            cache[{cur, clipboard}] = min(n1, n2);
            return min(n1, n2);
        }
    }
};
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