Classic problem but the off-by-one logic when you find the first mismatch trips people up.
Use a two-pointer technique from both ends of the string, moving inward while characters match. When a mismatch occurs, check if skipping either the left or right character results in a palindrome, and return true if either does. This yields O(n) time and O(1) space.
Pro tip: Clarify edge cases upfront (empty string, single character, already palindrome) and discuss the trade-off between the optimal two-pointer approach and a simpler brute-force O(n^2) method. This shows you consider efficiency and real-world constraints.
Confirm that removing at most one character means zero or one removal is allowed. Discuss edge cases: empty string, single character, and strings that are already palindromes.
Mention that a brute-force solution would try removing each character and checking for palindrome, taking O(n^2) time. Then propose the optimal two-pointer approach with O(n) time and O(1) space.
Initialize left and right pointers at the ends. While left < right, if characters match, move both inward. If they don't match, check if the substring skipping left or skipping right is a palindrome. If either is, return true; otherwise, return false.
Write a helper function that checks if a substring (given start and end indices) is a palindrome using two pointers. Use this helper when a mismatch occurs to test the two possible removals.
State that the time complexity is O(n) because each character is visited at most twice, and space is O(1). Walk through examples like 'abca' (true) and 'abc' (false) to verify correctness.
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