In this Hackerrank Iterables and Iterators problem we have
Input Format
The input consists of three lines. The first line contains the integer, denoting the length of the list. The next line consists of space-separated lowercase English letters, denoting the elements of the list.
The third and the last line of input contains the integer, denoting the number of indices to be selected.
Output Format
Output a single line consisting of the probability that at least one of the indices selected contains the letter:”.
Note: The answer must be correct up to 3 decimal places.
Problem solution in Python 2 programming.
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
N = input()
letters = raw_input().strip().split()
K = input()
N_a = letters.count('a')
if N_a == 0:
print '0'
elif N - N_a < K:
print '1'
else:
num = nCr(N - N_a, K)
denum = nCr(N, K)
print (denum - num) * 1.0 / denum
Problem solution in Python 3 programming.
# Enter your code here. Read input from STDIN. Print output to STDOUT
from itertools import combinations
N = int(input())
L = input().split()
K = int(input())
C = list(combinations(L, K))
F = filter(lambda c: 'a' in c, C)
print("{0:.3}".format(len(list(F))/len(C)))
Problem solution in pypy programming.
from __future__ import division import itertools n = int(raw_input()) alphas = raw_input().split() k = int(raw_input()) checkSymbol = 'a' combinations = list(itertools.combinations(alphas,k)) filtered = [cb for cb in combinations if checkSymbol in cb] print len(filtered)/ len(combinations)
Problem solution in pypy3 programming.
from itertools import combinations
# Enter your code here. Read input from STDIN. Print output to STDOUT
n = int(input())
lst = [i for i in input().split()]
k = int(input())
a_indxs = {i for i in range(1, n + 1) if lst[i - 1] == 'a'}
a_num = 0
comb_len = 0
for comb in combinations(range(1, n + 1), k):
comb_len += 1
if set(comb) & a_indxs:
a_num += 1
print(a_num / comb_len)
