class Solution:
def isPalindrome(self, s: str) -> bool:
"""
Check if the given string 's' is a palindrome.
This function converts all uppercase letters to lowercase and ignores non-alphanumeric characters.
It then uses the two-pointer technique to compare characters from both ends of the string.
Parameters:
s (str): The input string consisting of printable ASCII characters.
Returns:
bool: True if 's' is a palindrome after processing, otherwise False.
"""
left = 0 # Pointer starting from the beginning of the string.
right = len(s) - 1 # Pointer starting from the end of the string.
while left < right:
# Skip non-alphanumeric characters from the left.
while left < right and not s[left].isalnum():
left += 1
# Skip non-alphanumeric characters from the right.
while left < right and not s[right].isalnum():
right -= 1
# Compare the characters at the pointers in lowercase form.
if s[left].lower() != s[right].lower():
return False # If mismatch found, it's not a palindrome.
left += 1 # Move the left pointer rightward.
right -= 1 # Move the right pointer leftward.
return True # If all characters match, the string is a palindrome.
Summary of Techniques and Approaches:
Two-Pointer Technique: By using pointers starting at opposite ends of the string and moving towards the center, we efficiently check for palindrome properties in O(n) time with O(1) extra space.
In-Place Filtering: Instead of creating an extra string, we skip non-alphanumeric characters during the traversal, thereby saving memory and processing time.
Case Insensitivity Handling: Converting characters to lowercase during comparison makes the algorithm robust against differences in character case.
General Application: This approach is applicable to problems involving symmetric comparisons, in-place array modifications, or filtering elements while traversing a sequence. Look out for problems that require processing only parts of data that meet certain conditions (e.g., valid characters), and consider using pointer-based techniques for optimal performance.