<p>Given an&nbsp;integer <code>n</code>, <em>return a string with <code>n</code>&nbsp;characters such that each character in such string occurs <strong>an odd number of times</strong></em>.</p>

<p>The returned string must contain only lowercase English letters. If there are multiples valid strings, return <strong>any</strong> of them. &nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> &quot;pppz&quot;
<strong>Explanation:</strong> &quot;pppz&quot; is a valid string since the character &#39;p&#39; occurs three times and the character &#39;z&#39; occurs once. Note that there are many other valid strings such as &quot;ohhh&quot; and &quot;love&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> &quot;xy&quot;
<strong>Explanation:</strong> &quot;xy&quot; is a valid string since the characters &#39;x&#39; and &#39;y&#39; occur once. Note that there are many other valid strings such as &quot;ag&quot; and &quot;ur&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 7
<strong>Output:</strong> &quot;holasss&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 500</code></li>
</ul>

class Solution:
    def generateTheString(self, n: int) -> str:
        if n%2==0:
            return "a"*(n-1)+"b"
        else:
            return "a"*n
16300