Palindrome Number + Next-Greater Palindrome

Two-part HE coding round. First: detect whether an integer is a palindrome without converting to a string. Second: find the smallest palindrome strictly greater than the input. Requirements Part 1. isPalindrome(n: int) -> bool. Return True when the decimal representation of n reads the same forwards and backwards. Solve without converting to string. Negative numbers are not palindromes by convention. Part 2. nextPalindrome(n: int) -> int. Return the smallest palindrome strictly greater than n. Examples: 132 -> 141 99 -> 101 1221 -> 1331 Notes Part 1. Reverse the lower half of the integer and compare to the upper half. Pop the last digit (n % 10), build the reversed value, divide n by 10, stop when the reversed value is ≥ the remaining n. For odd-length numbers the middle digit lives on one side and you drop it before comparing. O(log₁₀ n) time, O(1) space. Part 2. Operate on the digit array: Mirror the left half onto the right (e.g. 12_3_45 → 12_3_21). If this mirrored value is already > n, return it. Otherwise, increment the center digit (or the middle pair, for even length), propagating carry into the left half, then mirror again. Edge case: the increment carries past the most significant digit (999 → 1001). Detect by length change and emit 10…01 with the right number of zeros. The full algorithm is O(L) in the number of digits and pure array manipulation. The two-step "mirror, then increment-if-needed" framing avoids special-casing each digit length. Preparation Pre-write Part 1 in five lines using the half-reverse trick. Drill on 1221, 121, 1, 0, -1, and 1000021. Practice Part 2 on paper for odd and even lengths and for the 999 → 1001 carry-over case. Watch the interviewer hint — some are willing to accept the string conversion for Part 1 and only push integer-math on Part 2; surface the integer-math approach unprompted to signal awareness.