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  2. Python
  3. Sets
  4. No Idea!
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No Idea!

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  • ccjav
     + 99 comments

    List comprehensions can do this easily, esp when you leverage that True == 1 and False == 0:

    n, m = raw_input().split()

sc_ar = raw_input().split()

A = set(raw_input().split())
B = set(raw_input().split())
print sum([(i in A) - (i in B) for i in sc_ar])

  • Anachor
     + 7 comments

    Instead of using multisets, its enough to use sets on A and B since set lookup is O(1) in python (Awesome, right?)

  • marinskiy
     + 8 comments

    Here is Python 3 solution from my HackerrankPractice repository:

    _ = input()
array = input().split()
like = set(input().split())
dislike = set(input().split())
print(sum((i in like) - (i in dislike) for i in array))

    Feel free to ask if you have any questions :)

  • jEeTu_pYthOn
     + 9 comments

    my simple python3 approach passes all test casses:

    a=input().split(' ')
l=([int(i) for i in input().split(' ')])
firstset=set([int(i) for i in input().split(' ')])
secondset=set([int(i) for i in input().split(' ')])
p=0
for i in l:
    if i in firstset:
        p+=1
    if i in secondset:
        p-=1
print(p)

  • jlrocha
     + 9 comments

    The problem is quickly resolved by using list comprehensions. This is my solution in Python 3:

    n, m = input().split()
arr = [int(x) for x in input().split()]
A = set([int(y) for y in input().split()])
B = set([int(z) for z in input().split()])
count = [0 + 1 if x in A else 0-1 if x in B else 0 + 0 for x in arr]
print(sum(count))
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