5. Longest Palindromic Substring
I didn't know how to do this one. I tried brute force to start, whereby I was comparing the first letter with the last and then moving the pointers. This didn't work and I ended up with multiple loops.
I googled and found the "Expand around the center" technique for determining palindromes. One way to find palindromes is to compare the first letter to the last letter, the second letter to the 2nd to last letter and so on. The other way is the "Expand around the center" technique.
def longestPalindrome(self, s: str) -> str:
if len(s) == 1:
# I stored the palindrome itself. Leetcode had an alternate solution where they store the position and
# max length of the anagram. At the end they return the substring based on that position and length.
# That approach is faster than my approach below as I'm getting the substring from each call to the
# expand_around_center function. And then I'm checking lengths on the substring. Just checking the maths
# is faster. This still works though.
longest_palindrome = s
for i in range(len(s)):
# find for odd sized, where the center is one character: "aba"
palindrome_odd = self.expand_around_center(s, i, i)
# find for even sized, where center is two characters: "abba"
palindrome_even = self.expand_around_center(s, i, i + 1)
temp_palindrome = palindrome_odd if len(palindrome_odd) > len(palindrome_even) else palindrome_even
longest_palindrome = longest_palindrome if len(longest_palindrome) >= len(temp_palindrome) \
def expand_around_center(self, s: str, left: int, right: int) -> str:
I returned a substring here. An alternative approach is to return the length of the substring
:param s: the string to search
:param left: index for left pointer to start at
:param right: index for right pointer to start at
:return: The substring
while left >= 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return s[left + 1: right]