<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;
        Node *left, *right;

        Node(int data) {
            val = data;
            left = NULL;
            right = NULL;
        }
    };

    int minStep = INT_MAX;

    void func(Node* node, int prevC, int steps, int n) {
        if (node->val > n) return;
        if (steps > minStep) return;

        if (node->val == n) {
            minStep = min(minStep, steps);
            return;
        }

        node->left = new Node(node->val + prevC);
        func(node->left, prevC, steps + 1, n);

        prevC = node->val;
        node->right = new Node(node->val + prevC);
        func(node->right, prevC, steps + 2, n);
    }

    int minSteps(int n) {
        if (n == 1) return 0;
        if (n == 2) return 2;

        Node* node = new Node(2);

        func(node, 1, 2, n);
        return minStep;
    }
};
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