Valid Palindrome

Problem

Given a string, return true if it reads the same forwards and backwards once you ignore non-alphanumeric characters and case.

Constraints

• 1 ≤ s.length ≤ 2·10⁵
• s may contain any printable ASCII

Approach — Two Pointers

This is a Two Pointers problem. The idea: walk two indices through the data together (or toward each other), turning an O(n²) pair search into one linear pass. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time, O(1) extra space.

Solution code

Python

class Solution:
    def isPalindrome(self, s):
        t = [c.lower() for c in s if c.isalnum()]
        return t == t[::-1]

Java

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

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.