<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 num) {
         if (num == 1) return s;
        int n = s.length();
        int it = 0;
          int col = (n / (2 * num - 2) + 1) * (num - 1);
        int move = num - 2;
        int c = 0;
        bool ans = true;

        vector<vector<char>> v(num, vector<char>(col, '#'));

        while (c != col) {
            // Fill downwards
            for (int j = 0; j < num; j++) {
                if (it >= n) {  // Corrected condition
                    ans = false;
                    break;
                }
                if (ans == false) break;
                v[j][c] = s[it];
                it++;
            }
            c++;

            // Fill diagonally upwards
            while (move != 0) {
                if (it >= n) {  // Corrected condition
                    ans = false;
                    break;
                }
                if (ans == false) break;
                v[move][c] = s[it];
                it++;
                c++;
                move--;
            }

            move = num - 2;  // Reset move after diagonal movement
        }

        string final = "";
        for (int i = 0; i < num; i++) {
            for (int j = 0; j < col; j++) {
                if (v[i][j] != '#') {
                    final += v[i][j];
                }
            }
        }

        return final;
    }
};

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