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  1. Prepare
  2. Python
  3. Itertools
  4. Iterables and Iterators
  5. Discussions

Iterables and Iterators

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932 Discussions

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  • codeandbeyondco
     + 0 comments

    This problem directing the old version documentation of itertools. It should be updated with the new version. https://docs.python.org/3.8/library/itertools.html

  • arushigupta1080
     + 0 comments

    i saw every answer but no where its mentioned that we need to iterate the list part or n times input

  • maluisamrok
     + 0 comments

    this work so good and fast!!!

    import itertools
N =int(input())
cadena =input().split()
k=int(input())

print(len(list(itertools.filterfalse(lambda a: 'a' not in a,itertools.combinations(cadena, k))))/(len (list(itertools.combinations(cadena, k)))))

  • neha_singh19aug
     + 0 comments
    from itertools import combinations

if __name__ == '__main__':
    length,l1,K = int(input()),input().split(),int(input())
    list_combos = list(combinations(l1, K))
    counter1 = 0
    for comb in list_combos:
            if 'a' in comb:
                counter1 += 1
    print(counter1 / len(list_combos))

  • Solspin
     + 0 comments

    I know this is for itertools practice, but without itertools the answer is quite clean. Using 1 minus the multiplicative probability of not choosing 'a'.

    N,s,n = int(input()),input(),int(input())
x = 1
b = N-s.count('a')
for i in range(n):
    x = x*((b-i)/(N-i)) 
print(1-x)