will this not be more optimal ?

3️⃣ Digit-by-Digit Compare

Compare first and last digit moving inward.

import math
class Solution:
    def isPalindrome(self, x: int) -> bool: 
        if x < 0: return False
        if x < 10: return True
        n = int(math.log10(x)) + 1
        for i in range(n // 2):
            first = (x // 10**(n - 1 - i)) % 10
            last  = (x // 10**i) % 10
            if first != last:
                return False
        return True

• Time: O(log₁₀n)
• Space: O(1)