HackerRank Python Solution - Sets - Set Mutations

• We have seen the applications of union, intersection, difference, and symmetric difference operations, but these operations do not make any changes or mutations to the set.

• We can use the following operations to create mutations to a set: .update() or |=

• Update the set by adding elements from an iterable/another set.

>>> H = set("Hacker")
>>> R = set("Rank")
>>> H.update(R)
>>> print H
set(['a', 'c', 'e', 'H', 'k', 'n', 'r', 'R'])

• .intersection_update() or &= Update the set by keeping only the elements found in it and an iterable/another set.

HackerRank Python Solution - Sets - Symmetric Difference

• If the inputs are given on one line separated by a character (the delimiter), use split() to get the separate values in the form of a list. The delimiter is space (ASCII 32) by default. To specify that comma is the delimiter, use string. split(','). For this challenge, and in general, on HackerRank, space will be the delimiter.

Usage:

>> a = raw_input()
5 4 3 2
>> lis = a.split()
>> print (lis)
['5', '4', '3', '2']

• If the list values are all integer types, use the map() method to convert all the strings to integers.

HackerRank Python Solution - Sets - .symmetric_difference() operation

.symmetric_difference():

• The .symmetric_difference() operator returns a set with all the elements that are in the set and the iterable but not both.

• Sometimes, a ^ operator is used in place of the .symmetric_difference() tool, but it only operates on the set of elements in the set.

• The set is immutable to the .symmetric_difference() operation (or ^ operation).

HackerRank Python Solution - Sets - .difference() operation

.difference():

• The tool .difference() returns a set with all the elements from the set that are not in an iterable. Sometimes the - operator is used in place of the .difference() tool, but it only operates on the set of elements in the set.

• Set is immutable to the .difference() operation (or the - operation).

HackerRank Python Solution - Sets - .intersection() operation

.intersection():

• The .intersection() operator returns the intersection of a set and the set of elements in an iterable. Sometimes, the & operator is used in place of the .intersection() operator, but it only operates on the set of elements in the set.

• The set is immutable to the .intersection() operation (or & operation).

HackerRank Python Solution - Sets - .union() Operation

.union():

• The .union() operator returns the union of a set and the set of elements in an iterable. Sometimes, the | operator is used in place of .union() operator, but it operates only on the set of elements in set.

• Set is immutable to the .union() operation (or | operation).

HackerRank Python Solution - Sets - .discard(), .remove() & .pop()

.remove(x):

• This operation removes element x from the set.

• If element x does not exist, it raises a KeyError.

• The .remove(x) operation returns None.

Example:

>>> s = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> s.remove(5)
>>> print s
set([1, 2, 3, 4, 6, 7, 8, 9])
>>> print s.remove(4)
None
>>> print s
set([1, 2, 3, 6, 7, 8, 9])
>>> s.remove(0)
KeyError: 0

HackerRank Python Solution - Sets - .add()

• If we want to add a single element to an existing set, we can use the .add() operation.

• It adds the element to the set and returns 'None'.

Example:

>>> s = set('HackerRank')
>>> s.add('H')
>>> print s
set(['a', 'c', 'e', 'H', 'k', 'n', 'r', 'R'])
>>> print s.add('HackerRank')
None
>>> print s
set(['a', 'c', 'e', 'HackerRank', 'H', 'k', 'n', 'r', 'R'])