<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>

#define MAX(a,b) ((a > b) ? a : b)
#define MIN(a,b) ((a < b) ? a : b)

class Solution {
public:
    string reverseStr(string s, int k) {
        string out;
        for (int i = 0; i < (s.size()/(2*k)) + 1; i++){
            int start = i * 2*k;
            int end = start + k;
            end = MIN(end,s.size());
            for (int j = (end-1); j >= start; j--){
                out += s[j];
            }
            for (int j = end; j < MIN((start + 2*k),s.size()); j++){
                out += s[j];
            }
        }
        return out;
    }
};
9100