Two pointers from both ends, compare characters, and when you hit a mismatch you get one free skip.
Use a two-pointer technique starting from both ends of the string. When characters differ, check if skipping either the left or right character results in a palindrome. If either check succeeds, return true; otherwise, return false.
Pro tip: Clarify that the solution runs in O(n) time because each character is visited at most twice, and mention that the space complexity is O(1) if you avoid creating substrings. Also, discuss edge cases like empty strings and strings with one character.
Restate the problem to ensure understanding: determine if a string can become a palindrome by removing at most one character. Ask about input constraints (e.g., length, character set) and expected output.
Explain that you'll use two pointers, one at the start and one at the end, moving inward while characters match. When a mismatch occurs, you'll check if skipping either character yields a palindrome.
Describe a helper function that checks if a substring (defined by indices) is a palindrome using two pointers. This avoids creating new strings and keeps space complexity O(1).
On mismatch, call the helper on the substring excluding the left character and on the substring excluding the right character. If either returns true, the answer is true; otherwise, false.
State that time complexity is O(n) because each character is processed at most twice, and space is O(1). Mention edge cases: empty string, single character, already palindrome, and strings requiring removal at different positions.
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