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Classic problem, you've probably seen it a hundred times.
Clarify edge cases like negative numbers and numbers ending in zero, then reverse only half of the integer to avoid overflow and achieve O(log n) time. Compare the reversed half with the remaining half, handling odd digit counts by ignoring the middle digit.
Pro tip: Mention that reversing only half the number prevents integer overflow and is more efficient; also note that negative numbers are not palindromes by definition, and numbers ending in zero (except zero itself) cannot be palindromes.
Ask about input range, negative numbers, and numbers ending in zero. State that negative numbers are not palindromes and that any positive number ending in zero (except 0) cannot be a palindrome.
Immediately return false for negative numbers and for numbers that end in zero but are not zero. Return true for single-digit numbers.
Iteratively extract the last digit of the original number and build the reversed half. Stop when the reversed half is greater than or equal to the remaining original half.
Check if the reversed half equals the remaining original half (for even digit counts) or if the reversed half divided by 10 equals the remaining half (for odd digit counts).
State that time complexity is O(log n) and space is O(1). Walk through examples like 121, 1221, 12321, 10, and -121 to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.