<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;  // Handle edge case

        int n = s.size();
        int cycleLen = 2 * numRows - 2;  // Calculate the length of a full cycle (zigzag)

        // Estimate the number of columns needed
        int numCols = (n / cycleLen) * (numRows - 1) + min(numRows, n % cycleLen);
        vector<vector<char>> mat(numRows, vector<char>(numCols, '0'));

        int row = 0, col = 0;

        for (int i = 0; i < n; i++) {
            mat[row][col] = s[i];

            if (row == 0) {
                row++;  // Start moving down
            } else if (row == numRows - 1) {
                row--;  // Start moving up
                col++;  // Move to the next column while going up
            } else {
                if (col % (numRows - 1) == 0) {
                    row++;  // Continue moving down
                } else {
                    row--;
                    col++;  // Move up diagonally
                }
            }
        }

        // Construct the final result by reading the matrix row by row
        string ans = "";
        for (int i = 0; i < numRows; i++) {
            for (int j = 0; j < numCols; j++) {
                if (mat[i][j] != '0') {
                    ans += mat[i][j];
                }
            }
        }

        return ans;
    }
};

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