<p>The string <code>&quot;PAYPALISHIRING&quot;</code> is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)</p>

<pre>
P   A   H   N
A P L S I I G
Y   I   R
</pre>

<p>And then read line by line: <code>&quot;PAHNAPLSIIGYIR&quot;</code></p>

<p>Write the code that will take a string and make this conversion given a number of rows:</p>

<pre>
string convert(string s, int numRows);
</pre>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;PAYPALISHIRING&quot;, numRows = 3
<strong>Output:</strong> &quot;PAHNAPLSIIGYIR&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;PAYPALISHIRING&quot;, numRows = 4
<strong>Output:</strong> &quot;PINALSIGYAHRPI&quot;
<strong>Explanation:</strong>
P     I    N
A   L S  I G
Y A   H R
P     I
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;A&quot;, numRows = 1
<strong>Output:</strong> &quot;A&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists of English letters (lower-case and upper-case), <code>&#39;,&#39;</code> and <code>&#39;.&#39;</code>.</li>
	<li><code>1 &lt;= numRows &lt;= 1000</code></li>
</ul>

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows <= 1){
            return s;
        }

        unordered_set<int> visited;
        queue<int> toVisit;

        int gap = (numRows - 1) * 2;

        for(int i = 0; i < s.length() + gap; i += gap){
            toVisit.push(i);
        }

        string result = "";

        while(!toVisit.empty()){
            int curr = toVisit.front();
            toVisit.pop();

            // cout << curr << ' ';

            if(visited.find(curr) != visited.end() || curr < 0 || curr > s.length() + gap){
                continue;
            }

            if(curr >= 0 && curr <= s.length() - 1){
                result += s[curr];
            }
            
            visited.insert(curr);

            toVisit.push(curr - 1);
            toVisit.push(curr + 1);
        }

        return result;
    }
};
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