Task
Given two sets A and B, compute 



(
A
∖
B
)
∪
(
B
∖
A
)
.


{\displaystyle (A\setminus B)\cup (B\setminus A).}



That is, enumerate the items that are in A or B but not both. This set is called the symmetric difference of A and B.

In other words: 



(
A
∪
B
)
∖
(
A
∩
B
)


{\displaystyle (A\cup B)\setminus (A\cap B)}

 (the set of items that are in at least one of A or B minus the set of items that are in both A and B).

Optionally, give the individual differences (



A
∖
B


{\displaystyle A\setminus B}

 and 



B
∖
A


{\displaystyle B\setminus A}

) as well.



Test cases
A = {John, Bob, Mary, Serena}
B = {Jim, Mary, John, Bob}



Notes
 If your code uses lists of items to represent sets then ensure duplicate items in lists are correctly handled. For example two lists representing sets of a = ["John", "Serena", "Bob", "Mary", "Serena"] and b = ["Jim", "Mary", "John", "Jim", "Bob"] should produce the result of just two strings: ["Serena", "Jim"], in any order.
 In the mathematical notation above A \ B gives the set of items in A that are not in B; A ∪ B gives the set of items in both A and B, (their union); and A ∩ B gives the set of items that are in both A and B (their intersection).


#include <iostream>
#include <set>
#include <algorithm>
#include <iterator>
#include <string>
 
using namespace std;
 
int main( ) {
   string setA[] = { "John", "Bob" , "Mary", "Serena" };
   string setB[] = { "Jim" , "Mary", "John", "Bob"  };
   set<string> 
       firstSet( setA , setA + 4 ),
       secondSet( setB , setB + 4 ),
       symdiff;
 
   set_symmetric_difference( firstSet.begin(), firstSet.end(),
                             secondSet.begin(), secondSet.end(),
                             inserter( symdiff, symdiff.end() ) );
 
   copy( symdiff.begin(), symdiff.end(), ostream_iterator<string>( cout , " " ) );
   cout << endl;
   return 0;
}