HackerRank] Symmetric Difference (저장용)

문제보기

Problem

Objective

Today, we're learning about a new data type: sets.

 

Concept

If the inputs are given on one line separated by a space character, use split() to get the separate values in the form of a list:

>> 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.

>> newlis = list(map(int, lis))

>> print (newlis)

[5, 4, 3, 2]

 

Sets are an unordered bag of unique values. A single set contains values of any immutable data type.

 

CREATING SETS

>> myset = {1, 2} # Directly assigning values to a set

>> myset = set() # Initializing a set

>> myset = set(['a', 'b']) # Creating a set from a list

>> myset {'a', 'b'}

 

MODIFYING SETS

Using the add() function:

>> myset.add('c')

>> myset {'a', 'c', 'b'}

>> myset.add('a') # As 'a' already exists in the set, nothing happens

>> myset.add((5, 4))

>> myset {'a', 'c', 'b', (5, 4)}

Using the update() function:

>> myset.update([1, 2, 3, 4]) # update() only works for iterable objects

>> myset

{'a', 1, 'c', 'b', 4, 2, (5, 4), 3}

>> myset.update({1, 7, 8})

>> myset

{'a', 1, 'c', 'b', 4, 7, 8, 2, (5, 4), 3}

>> myset.update({1, 6}, [5, 13])

>> myset

{'a', 1, 'c', 'b', 4, 5, 6, 7, 8, 2, (5, 4), 13, 3}

 

REMOVING ITEMS

Both the discard() and remove() functions take a single value as an argument and removes that value from the set. If that value is not present, discard() does nothing, but remove() will raise a KeyError exception.

>> myset.discard(10)

>> myset

{'a', 1, 'c', 'b', 4, 5, 7, 8, 2, 12, (5, 4), 13, 11, 3}

>> myset.remove(13)

>> myset

{'a', 1, 'c', 'b', 4, 5, 7, 8, 2, 12, (5, 4), 11, 3}

 

COMMON SET OPERATIONS 

Using union(), intersection() and difference() functions.

>> a = {2, 4, 5, 9}

>> b = {2, 4, 11, 12}

>> a.union(b) # Values which exist in a or b

{2, 4, 5, 9, 11, 12}

>> a.intersection(b) # Values which exist in a and b

{2, 4}

>> a.difference(b) # Values which exist in a but not in b

{9, 5}

The union() and intersection() functions are symmetric methods:

>> a.union(b) == b.union(a)

True

>> a.intersection(b) == b.intersection(a)

True

>> a.difference(b) == b.difference(a)

False

These other built-in data structures in Python are also useful.

 

문제부터 이해가 힘들었음....

 

 

 

M,arr=input(),set(map(int,input().split()))
N,arr1=input(),set(map(int,input().split()))
x=(arr.union(arr1)).difference((arr.intersection(arr1)))
print(*sorted(x,key=int),sep="\n")

m = int(input())
m_set = set(map(int, input().split()))
n = int(input())
n_set = set(map(int, input().split()))

print('\n'.join(map(str, sorted(m_set ^ n_set))))

if __name__ == '__main__' :
    ans_ = set()
    M = int(input())
    set_m_ = set(list(map(int, input().split()))[:M])
    N = int(input())
    set_n_ = set(list(map(int, input().split()))[:N])
    ans_.update(set_m_.difference(set_n_))
    ans_.update(set_n_.difference(set_m_))
    for i in sorted(ans_):
        print(i)