We can use the standard two-pointer approach that starts at the left and right of the string and move inwards. Whenever there is a mismatch, we can either exclude the character at the left or the right pointer. We then take the two remaining substrings and compare against its reversed and see if either one is a palindrome.

*- Yangshun*

```
class Solution(object):
def validPalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
# Time: O(n)
# Space: O(n)
left, right = 0, len(s) - 1
while left < right:
if s[left] != s[right]:
one, two = s[left:right], s[left + 1:right + 1]
return one == one[::-1] or two == two[::-1]
left, right = left + 1, right - 1
return True
```