<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:

    struct Node {
        int val;
        int parent;
        int ans;
    };

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

        Node * newNode = new Node();
        newNode->val= 2;
        newNode->parent = 1;
        newNode->ans = 2;
    
        queue<Node*> q;
        q.push(newNode);
        while(!q.empty()) {
            Node * f = q.front();
            q.pop();
            if(f->val == n) return f->ans;
            if(f->val * 2 <= n) {
                Node * newN = new Node();
                newN->val = f->val * 2;
                newN->parent = f->val;
                newN->ans = f->ans + 2;
                q.push(newN);
            }
            if(f->val + f->parent <= n) {
                Node * newN = new Node();
                newN->val = f->val + f->parent;
                newN->parent = f->parent;
                newN-> ans = f->ans + 1;
                q.push(newN);
            }
        }

        return 0;
    }

    int minSteps(int n) {
        int ans = bfs(n);
        return ans;
    }
};
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