Is Palindrome.md

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 * 10^5
• s consists only of printable ASCII characters.

valid_palindrome.java

class Solution {
    public boolean isPalindrome(String s) {
        
        if (s.length() == 0) return true;

        StringBuilder filtered = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (Character.isLetterOrDigit(s.charAt(i)))
                filtered.append(s.charAt(i));
        }

        return filtered.toString().equalsIgnoreCase(filtered.reverse().toString());
    }
}

valid_palindrome.kt

class Solution {
    fun isPalindrome(s: String): Boolean {
        if (s.isEmpty()) { return true }

        val filtered = StringBuilder()
        for (i in s.indices) {
            if (Character.isLetterOrDigit(s[i])) { filtered.append(s[i]) }
        }

        return filtered.reversed().toString().equals(filtered.toString(), ignoreCase = true) 
    }
}

valid_palindrome.py

# Solution with O(n) space complexity, O(n) time complexity
class Solution:
    def isPalindrome(self, s: str) -> bool:
        # filter input string into new string with lowercase alphanumbericals
        cleaned_s = "".join(e.lower() for e in s if e.isalnum())
        # iterate backwards and compare
        return cleaned_s[::] == cleaned_s[::-1]

valid_palindrome_pointers.py

# Solution using left & right pointers, with O(1) space complexity, O(logn) time complexity
class Solution:
	def isPalindromeWithPointers(s: str) -> bool:
		pointer_left, pointer_right = 0, len(s) - 1

		while pointer_left < pointer_right:
			# constant checks to ensure left not surpass right pointer
			while pointer_left < pointer_right and not s[pointer_left].isalnum():
				pointer_left += 1
			while pointer_left < pointer_right and not s[pointer_right].isalnum():
				pointer_right -= 1

			# compare
			if s[pointer_left].lower() != s[pointer_right].lower():
				return False

			# update pointers
			pointer_left, pointer_right = pointer_left + 1, pointer_right - 1

		return True