Iterables and Iterators in Python - HackerRank Solution
Problem :
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.
To read more about the functions in this module, check out their documentation here.
You are given a list of N lowercase English letters. For a given integer K, you can select any K indices (assume 1-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 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.
Constraints :
- 1 <= N <= 10
- 1 <= K < = N
All the letters in the list are lowercase English letters.
Sample Input :
4 a a c d 2
Sample Output :
0.8333
Explanation :
All possible unordered tuples of length 2 comprising of indices from 1 to 4
are:
(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)
Out of these 6 combinations, 5 of them contain either index 1 or index 2 which
are the indices that contain the letter 'a'. Hence, the answer is 5/6.
Solution :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | # Iterables and Iterators in Python - Hacker Rank Solution # Python 3 # Enter your code here. Read input from STDIN. Print output to STDOUT # Iterables and Iterators in Python - Hacker Rank Solution START from itertools import combinations N = int(input()) char = input().split() K = int(input()) count = 0; total = 0; for i in combinations(char,K): count += 'a' in i total += 1 print(count/total) # Iterables and Iterators in Python - Hacker Rank Solution END |
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