I'm only advancing the right side of the window by 1 or more places every iteration. So I was thinking my solution is O(n) but then I grow the window X amount of times depending on how many palindromes are found so O(2n). Can someone explain the big O for my solution, thanks!

```
class Solution {
public:
string longestPalindrome(string s) {
if(s.size() < 2)
{
return s;
}
int eos = s.size() - 1; // end of string
int r(0), l(0), loc(0), size(0); //right and left window, substring location, substring size
for(r = l + 1; r < s.size(); l = r++){
if(s[l] == s[r] || (l > 0 && s[l-1] == s[r])) //palindrom test for aa or aba
{
if(s[l] != s[r]) //assume aba type palindrom found
{
l--;
} else {
//find multiples in a row aaa, aaaaa, aaaaaa
while(r < eos && s[l] == s[r + 1]){
r++;
}
}
//find size of palindrom by growing window in both directions
while(l > 0 && r < eos && s[l-1] == s[r+1])
{
l--;
r++;
}
//if found palindrom bigger than current, record it's place
if(r-l+1 > size)
{
size = r-l+1;
loc = l;
}
//if found palindrom is bigger than 2, reset right to halfway
//between left and right side of the window
if(r-l > 2){
r = 1+r-((r-l)/2);
}
}
}
return size > 0 ? s.substr(loc,size) : s.substr(0,1);
}
};
```