My thought here is that you start from the outside working inwards. When you encounter a conflict you must decide whether to resolve it by deleting the left or right character. Trying both possibilities should be good enough. From that decision point you just keep comparing and if it fails that branch failed.
def check(s):
is_palindrome = lambda s: s == s[::-1]
if is_palindrome(s):
return True
d = [a==b for a, b in zip(s, s[::-1])].index(False)
return is_palindrome(s[d+1:len(s)-d]) or is_palindrome(s[d:-d-1])
I think the solution, logically, is to cut the string in two halves, duplicating the center character for odd-length strings (¿or can you ignore it a bit more?), computing edit distance between the front half and the reversed right half, allowing only deletions, and bailing out as soon as you know the answer is larger than one.
In real life, optimize away the “cut into two halves” and “reverse” parts, as they will be expensive.
Waiting with bated breath for the interview feedback... :-)
Godspeed!