<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

        k = numRows

        # rules
        # x = c % k
        # if x == 0: fill from s
        # if x > 0 and r = (k-1) -x: fill from s, else ""

        mat = [ [ ] for _ in range(numRows)]

        r = 0 
        s_pt = 0 

        while s_pt < len(s):

            c = len(mat[r])

            x = c % (k-1)
            ch = s[s_pt]

            if x == 0:
                mat[r].append(ch)
                s_pt += 1
            else:
                if r == k-x-1:
                    mat[r].append(ch)
                    s_pt += 1
                else:
                    mat[r].append("")

            r += 1
            r %= k 

        ans = ""
        for r in mat:
            ans += "".join(r)

        return ans

19200