HackerRank Iterables and Iterators solution in Python

The itertools module standardizes a core set of fast, memory efficient tools that are useful by themselves or in combination. Together, they form an iterator algebra making it possible to construct specialized tools succinctly and efficiently in pure Python.

You are given a list of N lowercase English letters. For a given integer K, you can select any K indices (assume -based indexing) with a uniform probability from the list.

Find the probability that at least one of the K indices selected will contain the letter: ‘a’.

Input Format

The input consists of three lines. The first line contains the integer N, denoting the length of the list. The next line consists of N space-separated lowercase English letters, denoting the elements of the list.

The third and the last line of input contains the integer K, denoting the number of indices to be selected.

Output Format

Output a single line consisting of the probability that at least one of the K indices selected contains the letter:’a’.

Note: The answer must be correct up to 3 decimal places.

HackerRank Iterables and Iterators solution in Python 2.

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

Iterables and Iterators solution in Python 3.

# 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)