Use a two-pointer technique starting from both ends of the string. When characters differ, check if the substring skipping the left character or the right character is a palindrome. If either is, return true; otherwise, return false.
Pro tip: Clarify edge cases upfront, such as empty strings, single-character strings, and strings that are already palindromes. Also, discuss the time and space complexity: O(n) time and O(1) space for the two-pointer approach.
Confirm that removing at most one character means zero or one removal is allowed. Discuss edge cases like empty string, single character, and strings with all identical characters.
Set left pointer at the start (0) and right pointer at the end (length-1) of the string.
While left < right, if characters at left and right are equal, move both pointers inward. If they differ, check if skipping the left character or skipping the right character results in a palindrome.
Write a helper function that checks if a substring is a palindrome using two pointers. Use it to verify the two possibilities when a mismatch occurs.
If the loop completes without mismatches, return true. If a mismatch is resolved by one removal, return true; otherwise, return false.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.