We can solve this without converting the number to a string by reversing the digits of the integer (or half of it to avoid overflow).

1. If x < 0 or (x % 10 == 0 and x != 0), return False.
2. Initialize rev = 0.
3. While x > rev:
   • rev = rev * 10 + x % 10.
   • x //= 10.
4. Return True if x == rev (even length) or x == rev // 10 (odd length).

• Time Complexity: O(log N).
• Space Complexity: O(1).

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