<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:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1:
            return s

        vertical = True
        numCols = math.ceil((len(s) / (numRows * 2)) * numRows)
        
        matrix = [["" for _ in range(numCols)] for _ in range(numRows)]  

        p, row, col = 0,0,0

        while p < len(s):
            print(row,col)
            matrix[row][col] = s[p]

            if vertical and row == numRows - 1:
                vertical = False
                row -= 1
                col += 1
                p += 1
                continue
            elif vertical:
                row += 1
            
            if not vertical and row == 0:
                vertical = True
                row += 1
            elif not vertical:
                row -= 1
                col += 1

            p += 1

        res = ""

        for i in range(numRows):
            for j in range(numCols):
                if not matrix[i][j] == "":
                    res += matrix[i][j]

        return res
        
20800