<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:
    bool visited[1001][1001] = {0};
    int minSteps(int n) {
        memset(visited, 0, sizeof(visited));
        queue<array<int, 3>> q;
        q.push({1, 0, 0});
        while (q.front()[0] != n) {
            auto curr = q.front();
            q.pop();
            visited[curr[0]][curr[1]] = true;
            if (curr[0] <= 1000 && !visited[curr[0]][curr[0]]) q.push({curr[0], curr[0], curr[2] + 1});
            if (curr[0] + curr[1] <= 1000 && !visited[curr[0] + curr[1]][curr[1]]) q.push({curr[0] + curr[1], curr[1], curr[2] + 1});
        }
        return q.front()[2];
    }
};
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