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[SNIPPET-22] HackerRank - Between Two Sets (Ugly Solution)
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#!/bin/python3 | |
import sys | |
from functools import reduce | |
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(' ')] | |
factors_of_b = [] | |
# TODO: learn about this algorithm | |
def factors(n): | |
return set(reduce(list.__add__, | |
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))) | |
for number in b: | |
factors_of_b.append(factors(number)) | |
temp = set() | |
for i in range(len(factors_of_b)-1): | |
temp = factors_of_b[i] & factors_of_b[i+1] | |
if (len(b) == 1): | |
temp = factors(b[0]) | |
# print(temp) | |
result = set() | |
for number in temp: | |
tempz = 0 | |
for other_number in a: | |
if number % other_number == 0: | |
tempz += 1 | |
if tempz == len(a): | |
result.add(number) | |
new_resultz = set() | |
for number in result: | |
tempz = 0 | |
for other_number in b: | |
if number in factors(other_number): | |
tempz +=1 | |
if tempz == len(b): | |
new_resultz.add(number) | |
print(len(new_resultz)) |
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