Valid Palindrome

Description

Question

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

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.

Clarification

paraphrase: Determine if provided string is a palindrome, taking into account only alphanumeric characters

• can the string be empty (yes, per example )
• are there uppercase and lowercase letters, and should they be counted as one letter

Solving

Approach 1 - Remove Non-alphanumeric , Two Pointers

class Solution {
    public boolean isPalindrome(String s) {
        char[] alphanumeric = s.toCharArray();
        int count = 0; 
 
        for (char c: alphanumeric) {
            if (Character.isAlphabetic(c) || Character.isDigit(c)) {
                alphanumeric[count] = Character.toLowerCase(c);
                count++;
            }
        }
 
        int left = 0;
        int right = count-1;
 
        while (left < right) {
            if (alphanumeric[left] != alphanumeric[right]) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

Time Complexity: O(n)
Space Complexity: O(n)

Approach 2 - Skip Non Alphanumeric, Two Pointers

class Solution {
    public boolean isPalindrome(String s) {
        char[] c = s.toCharArray();
        int left = 0;
        int right = s.length()-1;
 
        while (left < right) {
            char l = c[left];
            char r = c[right];
            if (!(Character.isAlphabetic(l) || Character.isDigit(l))) {
                left++;
                continue;
            }
            if (!(Character.isAlphabetic(r) || Character.isDigit(r))) {
                right--;
                continue;
            }
            if (Character.toLowerCase(l) != Character.toLowerCase(r)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

Time Complexity: O(n)
Space Complexity: O(n)