How to write a python function to return True when there's even number of 0 in a integer and re-CodePudding

I am asked to write a function that returns True if there's even number of 0 digits in a positive integer and return False otherwise (using recursion). As an attempt, I have written a function that counts the number of 0 in an integer. May I know how can I modify the program so that it will return True and False?

def valid(n):
    number = str(n)
    position = number.find("0")

    if "0" not in number:
        return 0

    return 1   valid(number[(position 1):])


print(valid(12340006))

CodePudding user response：

We can note that

[recursion base] Single zero is invalid, other single digit is valid
[recursion step] When we see non-zero digit, nothing should change (check for remaining part should give same result). When we see zero, we should invert the result (if it was odd count before, now it's even, and vice versa).

def is_valid(num):
    n, r = divmod(num, 10)
    if n == 0:
        return r != 0
    return (not is_valid(n) if r == 0 else is_valid(n))

CodePudding user response：

def valid(n):
    number = str(n)
    position = number.find("0")
    if "0" not in number:
        return 0
    return 1   valid(number[(position 1):])
print("True" if valid(12340006)%2 ==0 else "False")

CodePudding user response：

I believe the problem is the 0 returned from the if "0" not in number condition. You can try this way:

def valid(n):
    
    zeros = str(n).count("0")
    
    if zeros == 0:
        return False
    else:
        return zeros % 2 == 0

Which should work.

CodePudding user response：

You have a number, don't convert to string but rather use divmod and division by 10:

def valid(num, count=0):
    num, r = divmod(num, 10) # extract the last digit (r)
    if num == 0:             # we exhausted the number
        return count%2==0    # is there an even number of zeros?
    else:
        return valid(num, count=count int(r==0))

handling 0 -> False:

def valid(num, count=None):
    if count is None:
        count = int(num==0)
    num, r = divmod(num, 10)
    if num == 0:
        return count%2==0
    else:
        return valid(num, count=count int(r==0))

Example:

>>> valid(12340006)
False

>>> valid(10203)
True

>>> valid(0)
False

CodePudding user response：

If you are converting the number to a string using str, then you don't have to implement your own loop using recursion. You can count the number of '0' in the string using str.count, then test whether that number is odd or even using %.

def is_valid(n):
    s = str(n)
    nb_of_zeroes = s.count('0')
    return (nb_of_zeroes % 2 == 0)

If you want to implement the logic yourself, rather than using str and count, then you can loop on the digits of n with divmod(n, 10) to extract n (the unit digit) and n//10 (the number without its unit digit). For instance, divmod(9836, 10) is (983, 6).

def is_valid(n):
    if n == 0:
        return False  # special case for 0 which is a bit different from other numbers
    else:
        nb_of_zeroes = 0
        while n >= 10:
            n,u = divmod(n, 10)
            if u == 0:
                nb_of_zeroes  = 1
        return (nb_of_zeroes % 2 == 0)