# Solving Leetcode Interviews in Seconds with AI: Valid Palindrome


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "125" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers. Given a string s, return true if it is a palindrome, or false otherwise.   Example 1:  Input: s = "A man, a plan, a canal: Panama" Output: true Explanation: "amanaplanacanalpanama" is a palindrome.  Example 2:  Input: s = "race a car" Output: false Explanation: "raceacar" is not a palindrome.  Example 3:  Input: s = " " Output: true Explanation: s is an empty string "" after removing non-alphanumeric characters. Since an empty string reads the same forward and backward, it is a palindrome.    Constraints:  1 <= s.length <= 2 * 105 s consists only of printable ASCII characters.  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Two-Pointer Approach:** Use two pointers, one starting from the beginning and the other from the end of the processed string. Move the pointers towards the center, comparing characters at each step.
*   **Filtering Characters:** Efficiently filter out non-alphanumeric characters and convert uppercase letters to lowercase in a single pass.
*   **Early Exit:** If characters at the two pointers don't match, immediately return `False`.

*   **Runtime Complexity:** O(n), where n is the length of the input string.
*   **Storage Complexity:** O(1) (constant) - using only a few extra variables.

	
	# Code
	```python
	def isPalindrome(s: str) -> bool:
    """
    Given a string s, return true if it is a palindrome, or false otherwise.
    """
    left, right = 0, len(s) - 1

    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1

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

        left += 1
        right -= 1

    return True
	```