Valid Palindrome

Difficulty: Easy | Acceptance: 53.20% | Paid: No Topics: Two Pointers, String

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.

• Examples
• Constraints
• Two Pointers
• Filter and Reverse

Examples

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

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.

Two Pointers

Intuition Use two pointers starting from the beginning and end of the string. Move them towards each other, skipping non-alphanumeric characters, and compare the lowercased characters.

Steps

• Initialize left pointer at 0 and right pointer at s.length - 1.
• Loop while left < right.
• Increment left while it points to a non-alphanumeric character.
• Decrement right while it points to a non-alphanumeric character.
• Compare s[left] and s[right] (case-insensitive). If they differ, return false.
• Move both pointers inward.
• If the loop finishes, return true.

class Solution:
    def isPalindrome(self, s: str) -> bool:
        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

Complexity

• Time: O(n)
• Space: O(1)
• Notes: Optimal solution with constant extra space.

Filter and Reverse

Intuition Create a new string containing only alphanumeric characters converted to lowercase. Check if this new string is equal to its reverse.

Steps

• Iterate through the original string s.
• If a character is alphanumeric, convert it to lowercase and append it to a new string cleaned.
• Reverse the cleaned string.
• Compare cleaned with its reverse. Return true if equal, false otherwise.

class Solution:
    def isPalindrome(self, s: str) -> bool:
        cleaned = "".join(ch.lower() for ch in s if ch.isalnum())
        return cleaned == cleaned[::-1]

Complexity

• Time: O(n)
• Space: O(n)
• Notes: Uses extra space to store the cleaned string, but logic is very simple.