125. Valid Palindrome

July 2023 | Solving the easy valid palindrome LeetCode problem

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.


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.

Solution 1

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

        # remove all non-alphanumeric
        for char in s:
            if char.isalnum():
                new_string += char.lower()
        return new_string == new_string[::-1]

The solution uses extra memory by building a new string and then reversing it.

Solution 2: two pointers

Using constant memory (O(1)) using two pointers. Continuously increment the left and decrement the right pointer until they meet in the middle/pass each other

class Solution:
    def isPalindrome(self, s: str) -> bool:
        left, right = 0, len(s) - 1

        while left < right:
            # ensure that left does not pass right
            while left < right and not self.isAlphaNum(s[left]):
                left += 1
            while right > left and not self.isAlphaNum(s[right]):
                right -= 1

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

            left, right = left + 1, right - 1

        return True

    def isAlphaNum(self, c):
        return ord('A') <= ord(c) <= ord('Z') or ord('a') <= ord(c) <= ord('z') or ord('0') <= ord(c) <= ord('9')