Problem #125: Valid Palindrome

Intuition & Strategy Breakdown

Use two pointers (left and right) moving inward. Skip non-alphanumeric characters and compare lowercase characters for equality.

Python Solution

class Solution:
    def isPalindrome(self, s: str) -> bool:
        l, r = 0, len(s) - 1
        while l < r:
            while l < r and not s[l].isalnum(): l += 1
            while l < r and not s[r].isalnum(): r -= 1
            if s[l].lower() != s[r].lower():
                return False
            l, r = l + 1, r - 1
        return True

Java Solution

class Solution {
    public boolean isPalindrome(String s) {
        int l = 0, r = s.length() - 1;
        while (l < r) {
            while (l < r && !Character.isLetterOrDigit(s.charAt(l))) l++;
            while (l < r && !Character.isLetterOrDigit(s.charAt(r))) r--;
            if (Character.toLowerCase(s.charAt(l)) != Character.toLowerCase(s.charAt(r))) {
                return false;
            }
            l++; r--;
        }
        return true;
    }
}