<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 {
public:
    string reverseStr(string s, int k) {
        string t,ans;
        bool flag=false;
        for(int i=0;i<s.size()+1;i++){
            if(i!=0 && i%k==0){
                if(flag==false){
                    reverse(t.begin(),t.end());
                    flag=true;
                }
                else{
                    flag=false;
                }
                ans+=t;
                t.clear();
            }
            if(i<s.size())
                t.push_back(s[i]);
        }
        if(flag==false)
            reverse(t.begin(),t.end());
        ans+=t;
        return ans;
    }
};
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