What is the Big O run time of this solution:

def breakPalindrome(self, palindrome: str) -> str:

p = list(palindrome)

n = len(p)

if n == 1: return ""

for i in range(n):

if p[i] != "a":

p[i] = "a"

break

if p != p[::-1] : return "".join(p)

p = list(palindrome)

p[-1] = "b"

return "".join(p)