Check If A Number Is A Palindrome Without Changing It Into String

I'm having trouble with this problem that simply return True of False if a number n, is a palindrome. Note: wherever I have a ____ indicates where there is a blank that needs to b

Solution 1:

Here's the logic: (y * 10) + x % 10

defis_palindrome(n):
    x, y = n, 0
    f = lambda: (y * 10) + x % 10while x > 0:
        x, y = x//10 , f()
    return y == n

print(is_palindrome(123454321))
# Trueprint(is_palindrome(12))
# False

y*10 moves the current y to the left by 1 digit, and x%10 adds the last digit.

print(is_palindrome(235))
# False

Pre-iteration: x = 235, y = 0

First iteration: x = 23, y = 5

Second iteration: x = 2, y = 53

Third iteration: x = 0, y = 532

Solution 2:

Excellent solution, mate! Maybe one little suggestion. Your solution iterates over all n digits, but you have only to iterate n/2 digits. Moreover, you can handle negative values directly because they aren't palindromes.

def is_palindrome(x):
    if x < 0 or (x % 10 == 0 and x != 0):
        return False
    head, tail = x, 0
    whilehead > tail:
        head, tail = head // 10, tail * 10 + head % 10
    # When the length is an odd number, we can get rid of the middle digit by tail // 10returnhead == tail or head == tail // 10

Time complexity: O(log(n)) because we divided 10 in every iteration Space complexity: O(1)