<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;
        if(n == 3)  return 3;

        vector <bool> visited(n + 1 , false);

        queue <vector <int>> q;

        q.push({2 , 2 , 1});

        while(!q.empty()){
            int op = q.front()[0] , total = q.front()[1] , print = q.front()[2];
            q.pop();
            if(total == n)  return op;
            if(total + print <= n && !visited[total + print]){
                q.push({op + 1 , total + print , print});
            }
            if(total + total <= n && !visited[total + total]){
                q.push({op + 2 , total + total , total});
            }
        }

        return -1;
    }
};
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