Palindrome Number

Given an integer x, return true if its decimal digits read the same from left to right as from right to left, and false otherwise.

The minus sign counts as part of how the number is written, so a negative number is never a palindrome: -121 backwards would be 121-.

Try to solve it without converting the number to a string.

x % 10 peels off the last digit and integer division by 10 drops it. Repeating that builds the number backwards, one digit at a time.

You don't need the whole reversal, only enough to compare. Keep moving digits from the end of x onto a reversed number until the reversed part is at least as large as what's left. At that point you've consumed half the digits. For an even digit count the two halves must be equal; for an odd count the middle digit ends up on the reversed side, so drop it with one more division before comparing.

Stopping at the halfway point also means the reversed value can never overflow, which matters in languages with fixed-width integers.

def is_palindrome(x: int) -> bool:
    if x < 0 or (x % 10 == 0 and x != 0):
        return False
    reversed_half = 0
    while x > reversed_half:
        reversed_half = reversed_half * 10 + x % 10
        x //= 10
    return x == reversed_half or x == reversed_half // 10

function isPalindrome(x) {
  if (x < 0 || (x % 10 === 0 && x !== 0)) return false;
  let reversedHalf = 0;
  while (x > reversedHalf) {
    reversedHalf = reversedHalf * 10 + (x % 10);
    x = Math.floor(x / 10);
  }
  return x === reversedHalf || x === Math.floor(reversedHalf / 10);
}

Mention the string solution first and say it's fine for production code, then ask whether the interviewer wants the arithmetic version. Most do.

When you code the half reversal, trace one even-length number (1221) and one odd-length number (12321) out loud to show the two return conditions. Then name the two early exits (negatives, trailing zero) before you're asked about them.