Valid Palindrome II

https://leetcode.com/problems/valid-palindrome-ii/

Intuition:

• Set two iterators from the beginning and the end and make adjustments

Gotchas:

• the deletion might be needed from the either side of the iteration. So from the point of the mismatch, two verifications are needed
• slicing the char character may be tricky

Base cases:

• None

Solution:

class Solution:
    def validPalindrome(self, s: str) -> bool:
        i = 0
        j = len(s) - 1
        used = False
        while i <= j:
            if not s[i] == s[j]:
                return s[i:j] == s[i:j][::-1] or s[i+1:j+1] == s[i+1:j+1][::-1] # indexing
            i += 1
            j -= 1
        return True

Complexity Analysis:

• O(N)