Problem URL: Valid palindrome II

We will take two pointers, one at the beginning and one at the end. Then we start compare, if it doesn't match, then we try to skip the character in the left pointer and character in right pointer and check whether the rest is palindrome or not and return that. If the two pointers meet then it's a palindrome and we return that.

class Solution:
    def validPalindrome(self, s: str) -> bool:
        l, r = 0, len(s)-1
        while l<=r:
            if s[l] != s[r]:
                skipL, skipR = s[l+1:r+1], s[l:r]
                return skipL == skipL[::-1] or skipR == skipR[::-1]
            l += 1
            r -= 1
        return True

Time Complexity: O(n)
Space Complexity: O(n)