Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Note: This is easy if you convert to a string, then check if inputAsStr[::-1] == inputAsStr or something similar. However, I think that this is meaning for us to keep the integer type. So let’s solve it that way.

Solution

I find the best way to construct algorithms is to describe the scene, and by scene, I mean describe the items I need to achieve the output. This doesn’t have to include the functionality. For example:

Firstly, have a number n, and with n we need to work out if the it is the same number when reveresed. This means that we need a 2 vars. 1 to pop numbers off, and 1 to put those numbers on. What we will return is a bool if the var we add numbers to is the same as the input var.

def isPalindrome(self, n):
    """(x: int) -> bool"""
    x, y = n, 0

    return y == n

Okay, so now we have our basic scene, we need to introduce the logic of how we achieve the return. We need to go over the input x, and pop of the integer. As it is an integer, we can set x to be x//10. This will shift x to the right. We now need a function for y that both adds the end integer to y, and shifts y to the left for each new integer that is added to y. We can achieve this with a small lambda function. To see this working, lets have an input of 12321 and look at the output of x and y.

def isPalindrome(self, n):
    """(x: int) -> bool"""
    x, y = n, 0
    print(x, y)
    f = lambda: (y * 10) + x % 10
    while x > 0:
        x, y = x//10, f()
        print(x, y)
    return y == n

# 12321 0
# 1232 1
# 123 12
# 12 123
# 1 1232
# 0 12321

We start with our input and assign it to x and initilaise y to 0. We then divide out x by 10 giving 1232 and execute f(): (0 * 10) + 12321 % 10 = 1. On our next iteration x is 123, when f(): (1 * 10) + 1232 % 10 which is 12 etc etc.

Final

def isPalindrome(self, n):
    """(x: int) -> bool"""
    x, y = n, 0
    f = lambda: (y * 10) + x % 10
    while x > 0:
        x, y = x//10, f()
    return y == n