Run it
Both approaches on the same inputs, with the comparison counts printed — which is where the two-pointer version's short-circuit shows up. The last section is the delete-one follow-up.
The two solutions
Filter and compare:
O(n) time, O(n) space. Clear, short, and it allocates two extra sequences.
Two pointers:
O(n) time, O(1) space, and it short-circuits on the first mismatch rather than processing the whole string.
The lo < hi condition inside the skip loops matters: without it, a string of only punctuation runs the pointer past the end.
| | Filter and compare | Two pointers |
|---|
| Time | O(n) | O(n) |
| Space | O(n) | O(1) |
| Short-circuits | No | Yes |
| Lines | 2 | 10 |
Give the short version first, then offer the two-pointer version as the O(1)-space improvement. That sequence — correct answer, then optimisation — is what is being looked for.
Why isalnum and not arithmetic
The tempting optimisation is to compare character codes directly, or to lowercase by adding 32. Both are wrong on real input.
'0' is 48 and 'P' is 80, differing by 32 — the same offset as 'a' and 'A'. Any code that lowercases by adding 32 unconditionally will treat those as equal.
Non-ASCII characters break code arithmetic entirely. 'É'.isalnum() is True, and its code point is nowhere near the ASCII letters.
isalnum() and lower() handle both correctly, and they are C-level operations, so there is no performance argument for the arithmetic version.
Note that isalnum() is Unicode-aware: it returns True for digits and letters in any script, which is usually what is wanted and worth confirming.
The one-liner and its cost
s == s[::-1] is correct and reads well. It builds a full reversed copy first, so it is O(n) extra memory, and it always compares the entire string even when the first and last characters already disagree.
Cleaning first — clean = "".join(c.lower() for c in s if c.isalnum()) then comparing — is readable and costs a second full copy on top.
The two-pointer version
One pointer at each end. Advance each past anything that is not alphanumeric, compare the two characters case-folded, then step both inwards. No copy is made and the first mismatch ends it.
The subtlety is the guards: both inner skip loops need lo < hi in their condition, or a string of pure punctuation runs a pointer off the end. That is the bug this question is really probing.
The follow-up
"Now allow one character to be deleted." On a mismatch, try skipping the left character or the right one and check whether either remaining span is a plain palindrome. That is still O(n): you only get one chance to branch, so the two checks do not nest.
Edge cases
| Case | Result | Why |
|---|
"" | True | The loop does not run |
"a" | True | lo == hi immediately |
".,!" | True | Everything is skipped; the pointers meet |
"aa" | True | Both compared, equal |
"ab" | False | Immediate mismatch |
"0P" | False | Both alphanumeric, not equal |
| Mixed case | True if equal after lowering |
| Non-ASCII letters | Handled by isalnum and lower |
The punctuation-only case is the one that breaks naive two-pointer implementations, because the skip loops must be bounded by lo < hi rather than by the array length alone.
The variations
Palindrome with at most one deletion. The most common escalation. On a mismatch, try skipping the left character or the right one, and check whether either remaining substring is a palindrome:
Still O(n): the helper runs at most twice, and each run is bounded by the remaining length.
Longest palindromic substring. A substantially harder problem — expand around each of the 2n−1 centres for O(n²), or Manacher's algorithm for O(n).
Palindrome number. Reverse the digits arithmetically without converting to a string, which is the version asked when string operations are disallowed.
Palindrome linked list. Find the middle with fast and slow pointers, reverse the second half, compare. O(n) time and O(1) space, and it is a good demonstration of composing techniques.
Can these characters be rearranged into a palindrome? Count characters; at most one may have an odd count. A counting problem rather than a two-pointer one.
What it is testing
Do you handle the filtering requirement rather than assuming clean input?
Do you reach O(1) space? The filter-and-reverse version is correct and allocates; two pointers is the expected improvement.
Do you bound the skip loops? This is the one place the implementation genuinely breaks.
Do you avoid character arithmetic? The "0P" case is deliberately included in test suites to catch it.
Do you ask what counts as a character? Case, punctuation, whitespace and non-ASCII all need a decision.
The transferable pattern is the two-pointer scan from both ends with skipping — it appears in merging, partitioning, and any problem where two positions converge under a condition.
Recap in one screen
- Ignore case and non-alphanumeric characters unless told otherwise — ask first.
- Filter-and-compare is two lines and O(n) space; two pointers is O(1) space and short-circuits.
- Bound the skip loops with
lo < hi, or punctuation-only input runs past the end. - Use
isalnum() and lower(), never character arithmetic — '0' and 'P' differ by 32. - The standard escalations are "one deletion allowed" and "longest palindromic substring".