# Solving Leetcode Interviews in Seconds with AI: Valid Palindrome II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "680" 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
	> Given a string s, return true if the s can be palindrome after deleting at most one character from it.   Example 1:  Input: s = "aba" Output: true  Example 2:  Input: s = "abca" Output: true Explanation: You could delete the character 'c'.  Example 3:  Input: s = "abc" Output: false    Constraints:  1 <= s.length <= 105 s consists of lowercase English letters.  

	# Explanation
	*   **High-level approach:** Use two pointers, `left` and `right`, to iterate inwards from the start and end of the string. If `s[left]` and `s[right]` are not equal, it means one of them has to be removed to make the string a palindrome. Check if the substring from `left + 1` to `right` (inclusive) or the substring from `left` to `right - 1` (inclusive) is a palindrome.
*   **Palindrome check:** The palindrome check is done using a similar two-pointer approach.
*   **Early exit:** If at any point neither removing the left nor the right character results in a palindrome, return `False`.

*   **Runtime Complexity:** O(n), where n is the length of the input string.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def validPalindrome(s: str) -> bool:
    """
    Given a string s, return true if the s can be palindrome after deleting at most one character from it.

    Example 1:
    Input: s = "aba"
    Output: true

    Example 2:
    Input: s = "abca"
    Output: true
    Explanation: You could delete the character 'c'.

    Example 3:
    Input: s = "abc"
    Output: false

    Constraints:
    1 <= s.length <= 105
    s consists of lowercase English letters.
    """

    def is_palindrome(sub_s: str) -> bool:
        left, right = 0, len(sub_s) - 1
        while left < right:
            if sub_s[left] != sub_s[right]:
                return False
            left += 1
            right -= 1
        return True

    left, right = 0, len(s) - 1
    while left < right:
        if s[left] != s[right]:
            return is_palindrome(s[left + 1:right + 1]) or is_palindrome(s[left:right])
        left += 1
        right -= 1
    return True
	```