<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;
        }

        int n = int(s.size());
        int sections = ceil(n / (2 * numRows - 2.0));
        int numCols = sections * (numRows - 1);

        vector<vector<char>> matrix(numRows, vector<char>(numCols, ' '));

        int currRow = 0, currCol = 0;
        int currStringIndex = 0;

        // Iterate in zig-zag pattern on matrix and fill it with string
        // characters.
        while (currStringIndex < n) {
            // Move down.
            while (currRow < numRows && currStringIndex < n) {
                matrix[currRow][currCol] = s[currStringIndex];
                currRow++;
                currStringIndex++;
            }

            currRow -= 2;
            currCol++;

            // Move up (with moving right also).
            while (currRow > 0 && currCol < numCols && currStringIndex < n) {
                matrix[currRow][currCol] = s[currStringIndex];
                currRow--;
                currCol++;
                currStringIndex++;
            }
        }

        string answer;
        for (auto& row : matrix) {
            for (auto& character : row) {
                if (character != ' ') {
                    answer += character;
                }
            }
        }

        return answer;
    }
};
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