<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; 
        if(n == 2) return 2;

        queue<vector<int>> q;
        // previously copied, current text on the notepad, cost
        q.push({ 1, 2, 2 });

        while(!q.empty()){
            int m = q.size();
            for(int i = 0; i < m; i++){
                auto v = q.front();
                q.pop();
                int copy = v[0], text = v[1], cost = v[2];
                if(text == n) return cost;

                if(copy + text <= n) 
                    q.push({ copy, copy + text, 1 + cost});
                if(2 * text <= n)
                    q.push({ text, 2 * text, 2 + cost }); 
            }
        }

        return 0;
    }
};

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