Two pointer approach gets you most of the way there.
Use a two-pointer technique to compare characters from both ends, and when a mismatch occurs, check if skipping either the left or right character results in a palindrome. This yields an O(n) time and O(1) space solution.
Pro tip: Clarify edge cases upfront (empty string, single character, already palindrome) and mention that the solution can be implemented without extra space, which is often expected at Meta.
Confirm that 'at most one character' means zero or one removal, and that the string can be empty or have length 1. Ask if the string contains only lowercase letters or any characters.
Initialize left and right pointers at the start and end. Move them inward while characters match. If a mismatch occurs, check if the substring skipping left or skipping right is a palindrome.
Write a helper function that checks if a substring (given left and right indices) is a palindrome using two pointers. This avoids code duplication.
When a mismatch is found, return true if either skipping the left character or skipping the right character results in a palindrome. Otherwise, return false.
State that time complexity is O(n) because each character is visited at most twice, and space is O(1). Walk through test cases: 'aba' (true), 'abca' (true), 'abc' (false).
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