Sunday, July 28, 2019

Symmetric Difference - Hacker Rank Solution

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
Task
Given  sets of integers,  and , print their symmetric difference in ascending order. The term symmetric differenceindicates those values that exist in either  or  but do not exist in both.
Input Format
The first line of input contains an integer, .
The second line contains  space-separated integers.
The third line contains an integer, .
The fourth line contains  space-separated integers.
Output Format
Output the symmetric difference integers in ascending order, one per line.
Sample Input
4
2 4 5 9
4
2 4 11 12
Sample Output
5
9
11
12
Symmetric Difference - Hacker Rank Solution
# Enter your code here. Read input from STDIN. Print output to STDOU
M = raw_input();m = raw_input().split()
N = raw_input();n = raw_input().split()

print "\n".join(sorted(list(set(m) ^ set(n)),key=int))

5 comments:

  1. n = int(input())
    list1 = list(map(int,input().split()))
    n1 = int(input())
    list2 = list(map(int,input().split()))
    [print(i) for i in sorted(set(list1).difference(set(list2)).union(set(list2).difference(set(list1))))]

    ReplyDelete
    Replies
    1. correcting:)
      [print(i) for i in sorted(set(list1).difference(set(list2)).union(set(list2).difference(set(list1))))]

      Delete
  2. amazing use of for loop for printing the set

    ReplyDelete
  3. # Simple Solution You Can Try This !!!
    a = int(input())
    a1= set(map(int, input().split()))
    b = int(input())
    b1 = set(map(int, input().split()))
    adiff = a1.difference(b1)
    bdiff = b1.difference(a1)
    output = sorted(list(adiff.union(bdiff)))
    for i in output:
    print(i)

    ReplyDelete
  4. a = int(input())
    x= set(map(int, input().split()))
    b = int(input())
    y = set(map(int, input().split()))
    z = x.symmetric_difference(y)
    output = sorted(list(z))
    for i in output:
    print(i)

    ReplyDelete

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