<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) {
        // Edge case: if numRows is 1, no zigzagging is required
        if (numRows == 1) return s;

        // Initialize a 2D vector with numRows and enough columns for the string
        vector<vector<char>> vec(numRows, vector<char>(s.size(), '\0'));

        int i = 0, m = 0, n = 0;
        bool goingDown = true;  // Direction control

        // Fill the 2D vector with characters in a zigzag pattern
        while (i < s.size()) {
            vec[m][n] = s[i];
            i++;
            
            if (goingDown) {
                // Move down
                m++;
                if (m == numRows) {
                    m = numRows - 2;  // Go back up after reaching the bottom
                    n++;
                    goingDown = false;
                }
            } else {
                // Move up
                m--;
                n++;
                if (m == -1) {
                    m = 1;  // Start moving down after reaching the top
                    goingDown = true;
                }
            }
        }

        // Build the final string by concatenating non-empty characters
        string result;
        for (int row = 0; row < numRows; row++) {
            for (int col = 0; col < s.size(); col++) {
                if (vec[row][col] != '\0') {
                    result += vec[row][col];
                }
            }
        }

        return result;
    }
};
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