125. Valid Palindrome

January 18, 2024

https://leetcode.com/problems/valid-palindrome/description/

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.

code

class Solution {
    public boolean isPalindrome(String s) {
        int i = 0, j = s.length() - 1;
        while (i < j){
            if (!Character.isLetterOrDigit(s.charAt(i))){
                i++;
                continue;
            }
            if (!Character.isLetterOrDigit(s.charAt(j))){
                j--;
                continue;
            }
            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(j))){
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}

code

class Solution {
  public boolean isPalindrome(String s) {
    for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
      while (i < j && !Character.isLetterOrDigit(s.charAt(i))) {
        i++;
      }
      while (i < j && !Character.isLetterOrDigit(s.charAt(j))) {
        j--;
      }

      if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(j)))
        return false;
    }

    return true;
  }
}

code

class Solution:
    def isPalindrome(self, s: str) -> bool:
        i, j = 0, len(s)-1
        while i < j:
            while i < j and not s[i].isalnum():
                i += 1
            while i < j and not s[j].isalnum():
                j -= 1

            if s[i].lower() != s[j].lower():
                return False
            i += 1
            j -= 1
        return True

code

class Solution:
    def isPalindrome(self, s: str) -> bool:
        if s == '':
            return True
        final = re.sub(r'\W+', '', s).lower()
        return final == final[::-1]