I'm doing this HackerRank problem:

Consider two sets of positive integers, \$A=\{a_0, a_1, \ldots, a_{n-1}\}\$ and \$B=\{b_0, b_1, \ldots, b_{m-1}\}\$. We say that a positive integer, \$x\$, is between sets \$A\$ and \$B\$ if the following conditions are satisfied:

  1. All elements in \$A\$ are factors of \$x\$.
  2. \$x\$ is a factor of all elements in \$B\$.

Given \$A\$ and \$B\$, find and print the number of integers (i.e., possible \$x\$'s) that are between the two sets.

Assuming lists a and b as in the problem, the main idea is finding the LCM of list a and the GCD of list b, and then finding out how many multiples of the LCM divide the GCD perfectly without remainders. I ended up with this -

from fractions import gcd

def lcm(a, b):
    for i in xrange(max(a,b), a*b+1):
        if i%a==0 and i%b==0:
            l = i
    return l

def count(l, g):
    count = 1
    if l==g:
        return count
    elif g%l !=0 and l%g != 0:
        return 0
    elif l<g:
        for i in xrange(l, g, l):
            if g%i==0:
                count += 1
        return count
        for i in xrange(g, l, g):
            if l%i==0:
                count +=1 
        return count

if __name__ == '__main__':
    n,m = raw_input().strip().split(' ')
    n,m = [int(n),int(m)]
    a = map(int,raw_input().strip().split(' '))
    b = map(int,raw_input().strip().split(' '))
    l = reduce(lcm, a)
    g = reduce(gcd, b)
    print count(l, g)

But I pass only 7 of the 8 test cases, the last one getting terminated due to time out. I don't understand which part of my code would result in a long loop that might end up in a timeout.

P.S. I would also be very glad if any of you could point out any other inefficiencies or styling conventions in my code.

  • 1
    \$\begingroup\$ Given a plausible implementation of GCD (math, now), why are you searching for an LCM exhaustively? \$\endgroup\$ – greybeard Dec 13 '16 at 23:39
  • 1
    \$\begingroup\$ I'd look at two SO posts for GCD and LCM. \$\endgroup\$ – Peilonrayz Dec 14 '16 at 0:43
  • \$\begingroup\$ @Peilonrayz, thanks for that link. I think the method shown is more streamlined than mine but I still get the timeout for that particular test case. \$\endgroup\$ – Sidharth Samant Dec 14 '16 at 3:48
  1. It is a really bad idea to use a variable named l. I is hard to distinguish it from 1.
  2. All these functions using xrange are inefficient

the lcd function can be efficiently calculated by

from fractions import gcd
def lcd(a,b):

the count function is


where divisor_count is the number-of-divisors function

If the number n has the prime factor decomposition

$$n=p_1^{e_1}\cdot p_2^{e_2}\cdots p_k^{e_k}$$ then we have

$$ \text{divisor_count}(n)=(e_1+1)\cdot(e_2+1)\cdots(e_k+1)$$

This can be calculated in the following way:

def divisor_count(n):
    while i*i<=n:
        while n%i==0:
    if n>1:

This divisor_count function runs in $$O(\sqrt{n})$$ time, the xrange implementation uses $$O(n)$$ time.

  • \$\begingroup\$ I would suggest writing docstrings for lcd and divisor_count. \$\endgroup\$ – Gareth Rees Jan 10 '17 at 21:41

It turns out my count function was inefficient. Going through the discussion forms on the site, I found this which passed all the test cases -

def count(l, g):
    count = 0
    for i in xrange(l, g + 1, l):
        if g%i == 0:
            count += 1
    return count
  • \$\begingroup\$ reduce might work faster than loop for that case, try it. \$\endgroup\$ – Alex Dec 14 '16 at 9:04

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