<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) {
        vector<int> arr;
        priority_queue<vector<int>> q;
        int ans, op, c, l;

        arr.assign(3, 0);
        arr[2] = 1;
        q.push(arr);
        while (!q.empty()) {
            arr = q.top();
            q.pop();
            op = arr[0];
            c = arr[1];
            l = arr[2];
            if (l == n) {
                ans = op;
                break;
            }
            if (l > n) {
                continue;
            }
            arr[0]++;
            if (c > 0) {
                arr[2] += c;
                q.push(arr);
            }
            if (c != l) {
                arr[1] = l;
                arr[2] = l;
                q.push(arr);
            }
        }

        return ans;
    }
};

86840