Valid Palindrome

0Easy

Valid Palindrome

Given a string s, return true if it is a palindrome, considering only alphanumeric characters and ignoring case.

Rules:

Ignore everything that isn't a letter or digit (spaces, punctuation).
Comparisons are case-insensitive.

Key idea (Two Pointers):

Put left at the start and right at the end.
Skip non-alphanumeric characters under either pointer.
Compare the lowercased characters; on any mismatch return false.
Move both pointers inward until they meet.

Runs in O(n) time with O(1) extra space — no cleaned copy of the string is built.

Jnaneshwara Reddy Satti

Builder · ClearCode

Hi — I'm a 21yo CS grad from India. I built ClearCode while recovering from a surgery, because I genuinely believe DSA should be visual, free, and accessible to everyone. No VC funding. No team. Just me, my laptop, and a lot of chai. ☕

If ClearCode helped you understand even one concept — consider supporting it. Every rupee keeps the servers running and new videos coming.

Support via Razorpay

Found this helpful? Support ClearCode ☕ — keeps it free forever

Algorithm Visualizer

O(n)

Welcome

TEST CASE: []O(1)

clearcode.in

Execution Trace

1def isPalindrome(s):

2 left = 0

3 right = len(s) - 1

5 while left < right:

6 while left < right and not s[left].isalnum():

7 left += 1

8 while left < right and not s[right].isalnum():

9 right -= 1

11 if s[left].lower() != s[right].lower():

12 return False

14 left += 1

15 right -= 1

17 return True

0:00 / 7:57