<p>Given a string <code>s</code> and an integer <code>k</code>, reverse the first <code>k</code> characters for every <code>2k</code> characters counting from the start of the string.</p>

<p>If there are fewer than <code>k</code> characters left, reverse all of them. If there are less than <code>2k</code> but greater than or equal to <code>k</code> characters, then reverse the first <code>k</code> characters and leave the other as original.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "abcdefg", k = 2
<strong>Output:</strong> "bacdfeg"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "abcd", k = 2
<strong>Output:</strong> "bacd"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        length=len(s)
        bundles=length//k
        output=[]
        for bundle in range(bundles+1):
            if bundle%2==0:
                curr=s[bundle*k:(bundle+1)*k][::-1]
                output.append(curr)
            else:
                output.append(s[bundle*k:(bundle+1)*k])
        return "".join(output)
            
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