PalindromeNumber.py

# Question link - https://leetcode.com/problems/palindrome-number/description/?envType=study-plan-v2&envId=top-interview-150

class Solution(object):
    def isPalindrome(self, x):
        """
        :type x: int
        :rtype: bool
        """
        # Sol2: Inplace palindrome with O(x)
        
        # Negative 
        if x < 0: return False

        # Dividian - as div significant value for the left digit
        div = 1
        while x >= 10 * div:
            div *= 10

        while x:
            right = x % 10
            left = x // div

            if left != right:
                return False

            # Chop up tha digits Right - > (x % div) , left -> (x % div) // 10
            x = (x % div) // 10
            # Reduce the div upto 2 digit - > // 100
            div = div // 100

        return True 

        # Sol : This is the traditional solution ,with huge complexity
        # temp = x
        # rev = 0
        # while temp:
        #     r = temp % 10
        #     rev = rev * 10 + r
        #     temp = temp / 10

        # return rev == x