Valid Palindrome

Ignore punctuation and letter case, then check whether text reads the same both ways.

Start with the concrete trace below. It shows the state the algorithm must carry as it runs.

Problem trace

Valid Palindrome: Keep raw text in place, skip non-alphanumeric endpoints one at a time, and move both pointers only after a case-insensitive match.

Input and goalIgnore punctuation and letter case, then check whether text reads the same both ways.

Initialize at both raw endpointsFor "A, b a!", L=0 points to A and R=6 points to !.

LA0,1space2b3space4a5R!6

coveredRange[0..6]actioninitialize opposite-end pointers

Recognize it

Use opposite-end pointers when equality must hold symmetrically after ignoring characters, case, or other endpoint noise.

Keep true

Before each loop iteration, every accepted alphanumeric pair outside [left,right] matches case-insensitively; discarded outside characters are non-alphanumeric and cannot affect the answer.

Reuse it

When irrelevant data appears at either boundary, repair one endpoint at a time until both are comparable; only then apply the symmetric predicate and advance both.

Pattern: Two pointers at opposite ends.

Simple idea: Skip non-letter and non-number characters. Compare the next real character from each side, then move inward.

def valid_palindrome(text: str) -> bool:
   left, right = 0, len(text) - 1

   while left < right:
      if not text[left].isalnum():
         left += 1
      elif not text[right].isalnum():
         right -= 1
      elif text[left].lower() != text[right].lower():
         return False
      else:
         left += 1
         right -= 1
   return True

Cost: $O(n)$ time and $O(1)$ space.