Multiple of four

Note: It seems this particular page is indexed and ranked 1 in google with keyword some-interview-platofrm some-company. It’s unethical to search for solutions before you actually take the test. My blog is mainly about algorithms and it wasn’t my intention to help you scored 100% correct. It was purely my interest in algorithm that got me writing about it.

Damaged is done therefore I’m just gonna leave this post as it is with some keywords removed. I hope they can come out with other questions.

There were 2 questions.

1. Given a txt file find the number of occurences of a website url. Easy peasy.

2. $x*y=\text{multiple of 4}$. $x$ is given from STDIN and the answer is to find $y$ and do $z = 2(a) + b$. Where $a$ is the number of occurences of 4 and $b$ is $0$. Multiple of 4 must be in the format $4440000$ or $40$ or $440$. $404$ is wrong though. So you see 4 must come consecutively. But be sure to get it very fast.

STDIN
The first input 3 means 3 numbers to check.
The line follows are the number to check.

3
4
5
80

STDOUT
2
3
4

STDIN
5
62
67
20
63
81

STDOUT
30
66
3
36
162

Here’s the most optimal solution. The reason that makes it faster than looping through 0 until infinity while checking if it’s a multiple of 4 by calculation is that the code predetermined the value of four - 4, 40, 400, 440, 444, 4000, 4400, 4440, 4444 etc and just checking the modulus of it from there.

def checking(num):
    counter = 1
    while True:
        four = list("".join(str(e) for e in ['0' for x in range(counter)]))

        for x in range(1, counter + 1):
            print(x, counter)
            if x == 1:
                four[x - 1] = '4'
            else:
                four[: x] = ['4' for _ in range(x)]

            four_int = int("".join(four))

            if four_int % num == 0:
                count_0 = four.count('0')
                count_4 = four.count('4')

                return count_4, count_0

        counter += 1


N = int(input())

for x in range(N):
    num = int(input())
    a, b = checking(num)
    if a is None or b is None:
        print('Invalid')
    else:
        z = (2*a) + b
        print(z)