Started with the brute force and they let me run with it for a bit before nudging me toward something better.
Start by clarifying the problem: count all substrings that are palindromes, including duplicates. Then present an efficient solution using expand-around-center, which checks each possible center in O(n^2) time and O(1) space, and discuss trade-offs with other approaches like dynamic programming or Manacher's algorithm.
Pro tip: Mention that you would handle both odd and even length palindromes by expanding around each character and each gap between characters, and note that this approach avoids the O(n^2) space of DP while being simpler than Manacher's.
Confirm that we need to count all palindromic substrings, including single characters and duplicates, and discuss input constraints (e.g., string length, character set).
Select expand-around-center for its O(n^2) time and O(1) space, or mention dynamic programming (O(n^2) time and space) and Manacher's algorithm (O(n) time) as alternatives.
For each index i, expand around center i for odd-length palindromes and around center i and i+1 for even-length palindromes, incrementing a counter for each valid expansion.
State that the time complexity is O(n^2) because each center can expand up to O(n) times, and space complexity is O(1) as only a few variables are used.
Walk through a small example like 'aaa' to show the count (6) and verify the logic, and mention edge cases like empty string or single character.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.