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  4. Iterables and Iterators
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Iterables and Iterators

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  • 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)

  • ankitpatel003
     + 0 comments

    from itertools import combinations

    Input

    n = int(input()) # total number of letters letters = input().split() # list of lowercase letters k = int(input()) # number of indices to choose

    Generate all combinations of indices

    comb = list(combinations(letters, k))

    Count the combinations that contain at least one 'a'

    count = 0 for c in comb: if 'a' in c: count += 1

    Calculate the probability

    probability = count / len(comb) print(f"{probability:.4f}")

  • pithambar20
     + 0 comments
    from itertools import combinations

N = int(input())

englishLetters = input().split()

let = 'a'

K = int(input())

l1 = list(combinations(englishLetters,K))

# print(l1)
favOutcome = 0
totOutcome = len(l1)

for item in l1:
    if 'a' in item:
        favOutcome += 1    

    
            

prob = favOutcome/totOutcome


print(prob)