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  1. Practice
  2. Algorithms
  3. Implementation
  4. Between Two Sets
  5. Discussions

Between Two Sets

  • NicolasBi + 10 comments

    Thank you for your solution, you helped me produce this oversized Python 3 code :

    #!/bin/python3

import sys
from functools import reduce


def lcm(a: int, b: int) -> int:
    return a * b // gcd(a, b)


def lcm_list(lst: list) -> int:
    return reduce(lcm, lst)


def gcd(a: int, b: int) -> int:
    while a % b != 0:
        a, b = b, (a % b)
    return b


def gcd_list(lst: list) -> int:
    return reduce(gcd, lst)


def evenly_divides(number: int, divisor: int) -> bool:
    return (number % divisor) == 0


def get_input():
    n, m = input().strip().split()
    n, m = [int(n), int(m)]
    a = [int(a_temp) for a_temp in input().strip().split()]
    b = [int(b_temp) for b_temp in input().strip().split()]
    return n, m, a, b


def main():
    n, m, a, b = get_input()
    
    # Find LCM of all integers of a
    lcm_value = lcm_list(a)
    
    # Find GCD of all integers of b
    gcd_value = gcd_list(b)
    
    # Count the number of multiples of LCM that evenly divides the GCD.
    counter = 0
    multiple_of_lcm = lcm_value
    while multiple_of_lcm <= gcd_value:
        if evenly_divides(gcd_value, multiple_of_lcm):
            counter += 1
        multiple_of_lcm += lcm_value

    print(counter)

    
if __name__ == "__main__":
    main()
    • stregz + 1 comment

      perhaps easier to understand, but your organization is on point. Uses built-in gcd:

      #!/bin/python

import sys
from fractions import gcd

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(' '))

def LCM(a, b):
    return (a*b)//gcd(a,b)

lcm = reduce(LCM, A, 1)
gcd = reduce(gcd, B)

lcm_copy = lcm

count = 0
while lcm <= gcd:
    if(gcd % lcm) == 0:
        count += 1
    lcm = lcm + lcm_copy

print(count)
      • askarovdaulet + 5 comments

        Hmm, it seems the initializer in reduce(LCM,A,1) is not needed, since the behavior of reduce for single-element lists is just returning the element. For the sake of curiosity, here's a more packed version of your approach. Arguably, it does push the limits of readability :)

        import sys
from functools import reduce
from fractions import gcd

input()
a = map(int,input().strip().split())
b = map(int,input().strip().split())
lcm_a = reduce(lambda x,y: x*y//gcd(x,y), a)
gcd_b = reduce(gcd, b)
print(sum(1 for x in range(lcm_a,gcd_b+1,lcm_a) if gcd_b%x==0))
        • amelnychukoseen + 0 comments

          Damn this is nice.

        • mushthofa + 0 comments

          Nice

        • kanabos + 1 comment

          Not a big change but you can iterate less times:

          import sys
from functools import reduce
from fractions import gcd

input()
a = map(int,input().strip().split())
b = map(int,input().strip().split())
lcm_a = reduce(lambda x,y: x*y//gcd(x,y), a)
gcd_b = reduce(gcd, b)
gcd_b_copy = gcd_b
print(sum(1 for x in range(lcm_a,gcd_b+gcd_b_copy,lcm_a) if gcd_b%x==0))
          • xiaoliji + 1 comment

            1 is redundant. just 

            print(sum(gcd_b%x==0 for x in range(lcm_a,gcd_b+gcd_b_copy,lcm_a)))
        • dheerajpoonia29 + 0 comments

          Awesome

        • ssbreckenridge + 0 comments

          Note for python3.7:

          DeprecationWarning: fractions.gcd() is deprecated. Use math.gcd() instead.

    • P1zzAdr0hn3 + 2 comments
      #!/bin/python3

import sys


n,m = input().strip().split(' ')
n,m = [int(n),int(m)]
a = [int(a_temp) for a_temp in input().strip().split(' ')]
b = [int(b_temp) for b_temp in input().strip().split(' ')]
a.sort()
b.sort()
ausgabe = 0
for q in range(b[0] +1):
    if q >= a[-1]:
        for t in range(n):
            if q % a[t] != 0:
                break
            if t == n -1:
                for g in range(m):
                    if b[g] % q != 0:
                        break
                    if g == m -1:
                        ausgabe += 1
      
print(ausgabe)
      • AffineStructure + 2 comments

        3 for loops = O(n^3)

        • P1zzAdr0hn3 + 5 comments

          Yeah thats not nice... Your approach using lists and all() is much more elegant.

          #!/bin/python3

import sys

n,m = map(int, input().strip().split(' '))
a = list(map(int, input().split()))
b = list(map(int, input().split()))

ausgabe = 0
for q in range(max(a), min(b) +1):
    if all(q % arr == 0 for arr in a) and all(brr % q == 0 for brr in b):
        ausgabe += 1
    
print(ausgabe)
          • namanrocks94 + 1 comment

            amazing way, but would this be O^2 ??

            • P1zzAdr0hn3 + 1 comment

              i think its O(2n^2)

              • SalmanBhai + 1 comment

                O(2n^2) is supposed to be O(n^2). Constants aren't taken into account.

                • P1zzAdr0hn3 + 1 comment

                  Asymptomatic complexity... i think now i got it :D

                  • paul_schmeida + 1 comment

                    It's asymptotic actually :)

                    • P1zzAdr0hn3 + 0 comments

                      :D I knew it was something like that.

          • suryach750 + 0 comments

            can you test this i/p: 3 2 4 8 12 12 24

          • MatricksDecoder + 1 comment

            print(len([num for num in range(max(a),max(b)+1) if (all(num%i==0 for i in a)) and all(i%num==0 for i in b)]))

            • zad23 + 0 comments

              surely should be min(b)+1 not max(b)+1

          • ash_jo4444 + 0 comments

            Awesome solution !!

          • washimdas2013 + 0 comments

            Superb Dude!!

        • isabella_xy_sun + 2 comments

          Hi there. I'm a beginner in Python. I don't know too much about complexity. Below is my solution in Python 2. Is my way efficient? Thank you!

          #!/bin/python
import sys
n,m = map(int,raw_input().strip().split(' '))
a = map(int,raw_input().strip().split(' '))
b = map(int,raw_input().strip().split(' '))
final = 0
lower,upper = max(a),min(b)

for i in xrange(lower,upper+1):
    if sum(1 for k in a if i%k == 0) == n and sum(1 for k in b if k%i == 0) == m:
        final += 1
print final
          • AffineStructure + 1 comment

            Your solution is a bruteforce solution that checks each of the digits between the sets, therefore it is not efficient. The editorial shows a clever solution in O(sqrt(n)) solution.

            For the runtime of your code, you have two generator comprehensions, where each comprehension checks through each element inthe length of the list for every i in the range.

            If we were to compute what the runtime, it should be [(upper-lower) * (len(a) + len(b))] = O(n^2) #someone correct me if this is wrong

            • P1zzAdr0hn3 + 1 comment

              for, for would be O(n^2) but we have for, for and for O(2n^2)

              • AffineStructure + 1 comment

                With time complexity we drop out the constant term.

                https://en.wikipedia.org/wiki/Time_complexity

                "For example, if the time required by an algorithm on all inputs of size n is at most 5n^3 + 3n for any n (bigger than some n_0), the asymptotic time complexity is O(n^3)."

                • P1zzAdr0hn3 + 0 comments

                  ... you're right^^

          • chinmay_zare + 0 comments

            its most lucid and probably the best algorithm ! Thanks. it made me understand the concept.

      • skdubey + 0 comments

        your program has one error... it is not compile

    • __grj + 0 comments
      [deleted]
    • SebastianNielsen + 1 comment

      What is this syntax? 

      def gcd(a: int, b: int) -> int:

      I have never seen that before in python3, couldyou ellaborate on what it is for? At first i thought that gcd only accepted a and b to be an integer. But when I tried setting a to an float, no error accured.

      • What is it for?
      • aymanabuelela + 0 comments

        Function annotations https://www.python.org/dev/peps/pep-3107/

    • [deleted] + 3 comments

      Here is smaller code bro,

      def getTotalX(a, b):
    total = 0
    for x in list(range(a[-1], b[0] + 1)):
        hold = 0
        for v in a:
            if (x % v == 0): hold += 1
            else: hold = 0; break    
        
        for c in b:
            if (c % x == 0): hold += 1
            else: hold = 0; break       
        if (hold == len(a) + len(b)):
            total += 1   
    return total
      • anurag12345 + 0 comments

        Awesome bro

      • chandrakanthshy + 0 comments

        add a if condition above the 3rd for loop like:

        if hold > 0:

         for c in b:
 ....
 ....
      • gkjha009 + 0 comments

        Great,

        We not only need to solve this puzzle but at the same time think of time complexity :).

        We need to find numbers b/w two arrays, we can take max from a and min from b, then we can do calculation to find numbers.

        This is what i'm doing in m beloved JS.

        function getTotalX(a, b) { let total = 0; let maxA = Math.max(...a); let minB = Math.min(...b); let number = maxA;

        let allElementsAreMultiple = false;
let numberIsMultipleOfAll  = false;

while(number <= minB){
    console.log('aMax ', maxA,'bMin ', minB, 'number ', number);
    // Every element of array  must be a multiple of considerd number
    allElementsAreMultiple = a.every(ele => number%ele === 0 );
    numberIsMultipleOfAll  = b.every(ele => ele%number === 0 );
    if( allElementsAreMultiple && numberIsMultipleOfAll )
        total++;
    number++;
}

return total;

        }

        That makes it very easy for me.

    • jamesallenvinoy + 1 comment

      "lcm(a: int, b: int) -> int" , what is the concept in python behind this??

      • AffineStructure + 1 comment

        https://www.python.org/dev/peps/pep-3107/

        • AffineStructure + 0 comments

          Also this: https://stackoverflow.com/questions/14379753/what-does-mean-in-python-function-definitions

    • amitkumar10591 + 0 comments

      Will the solution return correct values when we use the below test case: 3 3 7 8 10 280 560 1120

      I think the solution above would give incorrect results.

    • suryabteja + 0 comments

      Cool code!! my code to find number of factors def nooffactors(z): print(sum(int(z)%i == 0 for i in range(1,z+1) )) return

    • martin89 + 0 comments

      This is far far more readable than prior solutions. I like it. SImilar to mine in c#

          static int getTotalX(int[] a, int[] b) 
{
    var lcm = LCM(a);
    var gcf = GCF(b);

    var candidate = lcm;
    var count = 0;
    while (candidate<=gcf)
    {
        count += gcf % candidate ==0 ? 1 :0;
        candidate += lcm;
    }
    return count;
}

static int LCM(int []x) => x.Aggregate(lCM);
static int GCF(int []x) => x.Aggregate(gCF);
static int gCF(int a, int b) => b == 0 ? a : gCF(b, a % b);  
static int lCM(int a, int b) => (a *b) / gCF(a, b);
    • nalin39 + 0 comments

      So thats what he meant(calculating the counter that way). Not sure, but some how I misunderstood "Integer being considered". Being a newbie here, just didn't notice this section. A lot has already been said and done, and I was going crazy over the tests. :-)