HackerRank Python Solution - Collections Topic - DefaultDict

The defaultdict tool is a container in the collections class of Python. It's similar to the usual dictionary (dict) container, but the only difference is that a defaultdict will have a default value if that key has not been set yet. If you didn't use a defaultdict you'd have to check to see if that key exists, and if it doesn't, set it to what you want.

For example:

 

from collections import defaultdict
d = defaultdict(list)
d['python'].append("awesome")
d['something-else'].append("not relevant")
d['python'].append("language")
for i in d.items():
    print i

HackerRank Python Solution - Itertools Topic - Maximize it!

• You are given a function f(X) = X². You are also given K lists. The i_th list consists of N_i elements.

• You have to pick one element from each list so that the value from the equation below is maximized:

• S = (f(X₁) + f(X₂) +....+ f(X_k)) % M

• X_i denotes the element picked from the i^th list. Find the maximized value S_max obtained.

• % denotes the modulo operator.

• Note that you need to take exactly one element from each list, not necessarily the largest element. You add the squares of the chosen elements and perform the modulo operation. The maximum value that you can obtain, will be the answer to the problem.

HackerRank Python Solution - Itertools Topic - iterables and iterators

• 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 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'.

HackerRank Python Solution - Itertools Topic - Compress the String!

• In this task, we would like for you to appreciate the usefulness of the groupby() function of itertools.

• 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.

• For a better understanding of the problem, check the explanation.

Input Format:

A single line of input consisting of the string S.

HackerRank Python Solution - Itertools Topic - itertools.combinations_with_replacement()

itertools.combinations_with_replacement(iterable, r):

• This tool returns r length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.

• Combinations are emitted in lexicographic sorted order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

Sample Code:

>>> from itertools import combinations_with_replacement
>>> 
>>> print list(combinations_with_replacement('12345',2))
[('1', '1'), ('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '2'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '3'), ('3', '4'), ('3', '5'), ('4', '4'), ('4', '5'), ('5', '5')]
>>> 
>>> A = [1,1,3,3,3]
>>> print list(combinations(A,2))
[(1, 1), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (3, 3), (3, 3), (3, 3)]

HackerRank Python Solution - Itertools Topic - itertools.combinations()

itertools.combinations(iterable, r):

• This tool returns the length subsequences of elements from the input iterable.

• Combinations are emitted in lexicographic sorted order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

Sample Code:

>>> from itertools import combinations
>>> 
>>> print list(combinations('12345',2))
[('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '4'), ('3', '5'), ('4', '5')]
>>> 
>>> A = [1,1,3,3,3]
>>> print list(combinations(A,4))
[(1, 1, 3, 3), (1, 1, 3, 3), (1, 1, 3, 3), (1, 3, 3, 3), (1, 3, 3, 3)]

HackerRank Python Solution - Itertools Topic - itertools.permutations()

itertools.permutations(iterable[, r]):

• This tool returns successive length permutations of elements in an iterable.

• If r is not specified or is None, then r defaults to the length of the iterable, and all possible full-length permutations are generated.

• Permutations are printed in a lexicographic sorted order. So, if the input iterable is sorted, the permutation tuples will be produced in sorted order.

Sample Code:

>>> from itertools import permutations
>>> print permutations(['1','2','3'])
<itertools.permutations object at 0x02A45210>
>>> 
>>> print list(permutations(['1','2','3']))
[('1', '2', '3'), ('1', '3', '2'), ('2', '1', '3'), ('2', '3', '1'), ('3', '1', '2'), ('3', '2', '1')]
>>> 
>>> print list(permutations(['1','2','3'],2))
[('1', '2'), ('1', '3'), ('2', '1'), ('2', '3'), ('3', '1'), ('3', '2')]
>>>
>>> print list(permutations('abc',3))
[('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'), ('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')]

HackerRank Python Solution - Itertools Topic - itertools.product()

itertools.product():

• This tool computes the cartesian product of input iterables.

• It is equivalent to nested for-loops.

• For example, product(A, B) returns the same as ((x,y) for x in A for y in B).

Sample Code:

>>> from itertools import product
>>>
>>> print list(product([1,2,3],repeat = 2))
[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]
>>>
>>> print list(product([1,2,3],[3,4]))
[(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)]
>>>
>>> A = [[1,2,3],[3,4,5]]
>>> print list(product(*A))
[(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)]
>>>
>>> B = [[1,2,3],[3,4,5],[7,8]]
>>> print list(product(*B))
[(1, 3, 7), (1, 3, 8), (1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 3, 7), (2, 3, 8), (2, 4, 7), (2, 4, 8), (2, 5, 7), (2, 5, 8), (3, 3, 7), (3, 3, 8), (3, 4, 7), (3, 4, 8), (3, 5, 7), (3, 5, 8)]