Valid Palindrome

1. Clarify the problem:

   • The problem requires determining whether a given string is a valid palindrome.
   • A palindrome is a string that reads the same forwards and backwards, ignoring non-alphanumeric characters and considering case insensitivity.
   • We need to return a boolean value indicating whether the string is a valid palindrome or not.

2. Analyze the problem:

   • Input: A string.
   • Output: A boolean value indicating whether the string is a valid palindrome.
   • Constraints:
     • The input string can contain alphanumeric characters and non-alphanumeric characters.
     • The input string can have leading or trailing spaces.
     • The comparison should be case-insensitive.

3. Design an algorithm:

   • To solve the problem, we can use a two-pointer approach.
   • We can initialize two pointers, one at the beginning of the string and the other at the end.
   • While the two pointers haven't crossed each other, we compare the characters at the pointers.
   • If the characters are alphanumeric and match, we move the pointers towards each other.
   • If the characters are not alphanumeric or they don't match, we return False, indicating that the string is not a valid palindrome.
   • If the pointers cross each other without encountering any non-matching characters, we return True, indicating that the string is a valid palindrome.

4. Explain your approach:

   • To solve the problem, we will use a two-pointer approach to check for palindrome validity.
   • We will define two pointers, left and right, initialized to the start and end of the string, respectively.
   • We will use a while loop to compare characters at the pointers while they haven't crossed each other.
   • Inside the loop, we will check if the characters at the pointers are alphanumeric and if they match (ignoring case).
   • If the characters are not alphanumeric or they don't match, we return False.
   • If the characters are alphanumeric and they match, we increment left and decrement right to continue comparing the next characters.
   • Finally, if the pointers cross each other without any non-matching characters, we return True.

5. Write clean and readable code:

python

def isPalindrome(s):
    left = 0
    right = 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 left < right and s[left].lower() != s[right].lower():
            return False

        left += 1
        right -= 1

    return True

6. Test your code:

   • Test Case 1:

     • Input: s = "A man, a plan, a canal: Panama"
     • Explanation: The input string is "A man, a plan, a canal: Panama". Ignoring non-alphanumeric characters and case, it reads the same forwards and backwards.
     • Expected output: True

   • Test Case 2:

     • Input: s = "race a car"
     • Explanation: The input string is "race a car". Ignoring non-alphanumeric characters and case, it does not read the same forwards and backwards.
     • Expected output: False

   • Test Case 3:

     • Input: s = "abccba"
     • Explanation: The input string is "abccba". Ignoring non-alphanumeric characters and case, it reads the same forwards and backwards.
     • Expected output: True

7. Optimize if necessary:

   • The algorithm has a time complexity of O(n), where n is the length of the input string. We iterate through the string once using the two-pointer approach.
   • The space complexity is O(1) since we only use a constant amount of space for the two pointers.

8. Handle error cases:

   • The code can handle empty strings or strings with only non-alphanumeric characters correctly.
   • The algorithm will return True for empty strings or strings with no alphanumeric characters since they can be considered valid palindromes.

9. Discuss complexity analysis:

   • The time complexity of the algorithm is O(n), where n is the length of the input string. We iterate through the string once using the two-pointer approach.
   • The space complexity is O(1) since we only use a constant amount of space for the two pointers.