Valid Palindrome - Leetcode 125 - 2 Pointers (Python)

[@GregHogg]

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[@Code_Creator123]

Great explanation brother , thanks From India 🎉🎉❤

[@GregHogg]

Glad to hear it! 🔥

[@vidhansaini9296]

s = ''.join(c for c in s if c.isalnum()) s = s.lower() if s == s[::-1]: return True return False can solve it like this as well without using pointer but space complexity is now o(n)

[@mikekertser5384]

def isPalindrome(word): return word.lower() == ''.join(reversed(word.lower()))

[@manishluckyreddy6710]

Can we solve the same problem using recursion? if yes, how can we do it. I'm trying to do so but I'm not getting it. Thanks in advance.

[@GregHogg]

Yes you could, although I think this solution is a lot more intuitive

[@praveenchandkakarla406]

@@GregHogg Exactly it's a clear explanation. Great work, Thanks!! Greg

[@KasthuriarachichigePrasanna]

great bro

[@technicallytechnical1]

explain this bro : class Solution(object): def isPalindrome(self, s): return re.sub(r'[^A-Za-z0-9]', '', s.lower())==re.sub(r'[^A-Za-z0-9]', '', s.lower()[::-1])

[@Redaxi]

re is the regex module, re.sub takes the regex pattern as a parameter which is "[^A-Za-z0-9]" here. This re.sub removes what is not in A-Z and a-z and 0-9. So basically it removes all that is non alphanumeric. then does s.lower, to make it all lowercase, on the right side it does the same thing but with [::-1] which is pythons slice op it takes start end step, start and end are empty so it takes the whole string and step is -1 so it goes in reverse, resulting in the reversed string, and if the reversed and normal modified strings are same then it returns True, i.e. string is a valid palindrome

[@technicallytechnical1]

@@Redaxi thanks but I knew it weeks prior 🫱🏽‍🫲🏼

[@roccococolombo2044]

Also: def isPalindrome(s: str): # Clean s str_cleaned = ''.join(char.lower() for char in s if char.isalpha()) # Compare forward to backward return str_cleaned == str_cleaned[::-1]