Use a two-pointer technique starting from both ends, moving inward while characters match. When a mismatch occurs, check if the remaining substring is a palindrome by skipping either the left or right character, and if either works, the string can become a palindrome with at most one removal. This runs in O(n) time because each character is visited at most twice.
Pro tip: Clarify that 'removing at most one character' means you can also remove zero characters, so a string that is already a palindrome should return true. Also, mention that the check for the remaining substring after a mismatch must be done in O(n) to maintain overall O(n) time, and you can achieve this by using a helper function that checks if a substring is a palindrome.
Confirm that the string consists of lowercase letters, and that removing at most one character means zero or one removal is allowed. Also, confirm that an empty string or a single-character string is considered a palindrome.
Set left pointer at the start (0) and right pointer at the end (n-1) of the string. Move them inward while the characters at left and right are equal.
When a mismatch occurs, check if the substring from left+1 to right is a palindrome, or if the substring from left to right-1 is a palindrome. If either is true, return true; otherwise, return false.
Write a helper function that checks if a substring is a palindrome using two pointers, running in O(n) time. Ensure that this helper is called at most twice, so overall time remains O(n).
Explain that the two-pointer traversal takes O(n) time, and the helper function also takes O(n) time, but since it's called at most twice, the overall time complexity is O(n). Space complexity is O(1) as no extra data structures are used.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.