<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 = s.size();
        
        string ans = "";
        
        int k = 0;
        vector<vector<char>> temp;
        for(int i = 0; i < numRows*2 - 2; i++) {
            temp.push_back({});
            for(int j = i; j < n; j += (numRows - 2)*2+2){
                temp[i].push_back(s[j]);
            }
        }
        
        for(int i = 0; i < temp.size(); i++) {
            for(int j = 0; j < temp[i].size(); j++) {
                cout << temp[i][j] << " ";
            }
            cout << "\n";
        }
        
        for(int i = 0; i < temp[0].size(); i++) {
            ans += temp[0][i];
        }
        
        int l = 1, r = temp.size() - 1;
        while(l < r) {
            int len = max(temp[l].size(), temp[r].size());
            for(int i = 0; i < len; i++) {
                ans += temp[l][i];
                if(i < temp[r].size()) {
                    ans += temp[r][i];
                }
            }
            l++;
            r--;
        }
        if(temp.size() > 1) {
            for(int i = 0; i < temp[l].size(); i++) {
                ans += temp[l][i];
            }
        }
        
        
        return ans;
    }
};
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