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HackerRank Python Challenges

In this article I show how I solved HackerRank Python challenges. The main purpose of the content is to present my acquired experiences primarily for learning purposes.

Other categories:

Introduction
Basic Data Types
Strings
Sets
Itertools
Collections
Built-Ins
Numpy
Others

Itertools

itertools.product()

Task:

You are given a two lists A and B. Your task is to compute their cartesian product AXB.

Solution:

a = input().split()
b = input().split()
a1 = []
b1 = []

for i in a:
    a1.append(int(i))

for i in b:
    b1.append(int(i))

prod = []

prod = [(x, y) for x in a1 for y in b1]

str_prod = ''
for i in prod:
    str_prod += str(i) + ' '

print(str_prod)

itertools.permutations()

Task:

You are given a string S. Your task is to print all possible permutations of size k of the string in lexicographic sorted order.

Solution:

from itertools import permutations

s = input().split()

text = s[0]
length = s[1]

#print(permutations(text, int(length)))
l = [''.join(p) for p in permutations(text, int(length))]
l.sort()

for i in l:
    print(i)

itertools.combinations()

Task:

You are given a string S. Your task is to print all possible combinations, up to size k, of the string in lexicographic sorted order.

Solution:

from itertools import combinations

s = input()
arr = s.split()

sList = []
sList[:0] = arr[0]
sList.sort()

for i in range(1, int(arr[1]) + 1):
    for j in list(combinations(sList, i)):
        line = ''
        for k in j:
            line += k
        print(line)

itertools.combinations_with_replacement()

Task:

You are given a string S. Your task is to print all possible size k replacement combinations of the string in lexicographic sorted order.

Solution:

from itertools import combinations_with_replacement

s = input()
s_split = s.split()

letter_list = []
letter_list[:0] = s_split[0]
letter_list.sort()

for i in [''.join(x) for x in list(combinations_with_replacement(letter_list, int(s_split[1])))]:
    print(i)

Compress the String!

Task:

In this task, we would like for you to appreciate the usefulness of the groupby() function of itertools . To read more about this function, Check this out . You are given a string S. Suppose a character 'c' occurs consecutively X times in the string. Replace these consecutive occurrences of the character 'c' with (X, c) in the string.

Solution:

from itertools import groupby

s = input()

s_list = []
s_list[:0] = s
result_list = ''

for k, g in groupby(s_list):
    result_list += '(' + str(len(list(g))) + ', '
    result_list += str(k) + ') '
    
print(result_list)

Iterables and Iterators

Task:

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'.

Solution:

import itertools

n = input()
letters = input()
k = int(input())

combinations = 0
contains = 0

for i in itertools.combinations(letters.split(), k):
    combinations += 1
    for j in i:
        if j == 'a':
            contains += 1
            break

print(contains / combinations)