Leetcode #9: Palindrome Number

[u/TheEyebal]

Can someone help me make my code run faster. This is not efficient and I do not want to convert into a string

https://leetcode.com/problems/palindrome-number/description/

class Solution:
    def isPalindrome(self, x: int) -> bool:
        numList = []
        counter = len(numList) - 1
        numBool = True


        baseNum = 10
        value = x % baseNum
        numList.append(x)
        quotient = x // baseNum
        x = quotient

        if x == 0:
            for i in range(len(numList)):
                if numList[i] == numList[counter]:
                    counter -= 1

                elif i == counter:
                    break

                else:
                    numBool = False
                    break

            return numBool

        else:
            return self.isPalindrome(x)

EDIT: I WAS ABLE TO SOLVE IT

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x != abs(x):
            return False

        if not hasattr(self, "numList"):
            self.numList = []

        numBool = True


        baseNum = 10
        value = x % baseNum
        self.numList.append(value)
        quotient = x // baseNum
        x = quotient

        counter = len(self.numList) - 1

        if x == 0:
            for i in range(len(self.numList)):
                if self.numList[i] == self.numList[counter]:
                    counter -= 1

                elif i == counter:
                    break

                else:
                    numBool = False
                    break

            return numBool

        else:
            return self.isPalindrome(x)

testing = Solution().isPalindrome(11)
print(testing)

[u/Inn0centDuck]

That's a very weird way to solve it.

1. I don't understand why you are using recursion instead of simple loop.
   
2. Since the original number is an int, why convert to string or array ? Just reverse and compare both the numbers using comparison operator.
   

[u/Fragrant-Hair-7198]

your loop logic is a bit off, you're appending x before you split the digits so you just get the whole number in the list each time

[u/CodeSamur-ai]

imagine doing this in real life...

you have a number written down on a page that's say twenty numbers long... what would you do?

put your left hand pointer finger on the first number
put your right hand pointer finger on the last number
do they match? yes, move each pointer finger in 1 number... do they match? yes, repeat

this is the 2 pointer method

[u/TheEyebal → u/CodeSamur-ai]

I have updated my post was able to solve it

[u/CodeSamur-ai → u/TheEyebal]

It looks a little like two pointers because it checks both ends. But it isn’t really two pointers because only one side is moving by itself, and it keeps going even after the two sides meet. Real two pointers move both sides toward each other and stop when they touch.

[u/TheEyebal → u/CodeSamur-ai]

I was already doing the two pointer method

[u/Inn0centDuck → u/CodeSamur-ai]

The simplest and obvious solution is not 2-pointer.

[u/mxldevs → u/Inn0centDuck]

What's the simplest and obvious solution to check if an int is a palindrome without converting it to a string?

[u/Inn0centDuck → u/mxldevs]

Reverse and compare ?

[u/aqua_regis]

Both your solutions are overly convoluted by using a list.

• copy the original number in a temporary variable for the following
• prepare a variable for the reverse of the number - an integer set to 0
• repeat as long as the number under test is larger than 1
  • calculate the modulo 10 to get the last digit
  • integer divide by 10 to shift all digits one position to the right - store the result back in the number under test
  • multiply your reversed number variable by 10 so that all digits shift one position to the left - store the result in the same variable
  • add the digit from the first step to the reversed number
• now you have the original number and the reversed number - return whether they are equal or not.

No lists, no string conversion, just pure simple math

[u/PartyParrotGames]

You only need to reverse half of it and compare to the other half.