print non common elements from two arrays - c++

Question

I am trying to print the non common elements between two arrays. For example,
if array1 = { 1, 3, 5} and array2 = {1, 2, 4, 5}, my output should be {2, 3, 4}.
I have tried something here. But it prints only 3.
What am I doing wrong?
#include<iostream>
using namespace std;
int main()
{
int a[] = { 1, 3, 5 };
int b[] = { 1, 2, 4, 5 };
bool contains = false;
int result[10];
int r = 0;
int x;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 4; j++) {
if (a[i] == b[j]) {
contains = true;
break;
}
}
if (!contains) {
result[r]=a[i];
++r;
}
else{
contains = false;
}
}
for (x = 0; x < r; x++)
{
cout<< result[x]<<"\n";
}
return 0;
}

Answer 1

You did not add the unique values from b to result. But this is c++. I recommend to use std::vector, std::sort, std::set_symmetric_difference and std::back_inserter.
#include <iostream> // std::cout
#include <algorithm> // std::set_symmetric_difference, std::sort
#include <iterator> // std::back_inserter
#include <vector> // std::vector
int main()
{
std::vector< int > a = { 1, 3, 5 };
std::vector< int > b = { 1, 2, 4, 5 };
std::sort( a.begin(), a.end() );
std::sort( b.begin(), b.end() );
std::vector< int > result;
std::set_symmetric_difference( a.begin(), a.end(), b.begin(), b.end(), std::back_inserter(result) );
for ( int x : result )
std::cout << x << "\n";
return 0;
}
Live Example
As suggested by #ChristianHackl in the comment below, it is also possible to use arrays for a and b and using the free-standing functions std::begin and std::end:
#include <iostream> // std::cout
#include <algorithm> // std::set_symmetric_difference, std::sort
#include <iterator> // std::back_inserter
#include <vector> // std::vector, std::begin, std::end
int main()
{
int a[]{ 1, 3, 5 };
int b[]{ 1, 2, 4, 5 };
std::sort( std::begin(a), std::end(a) );
std::sort( std::begin(b), std::end(b) );
std::vector< int > result;
std::set_symmetric_difference( std::begin(a), std::end(a), std::begin(b), std::end(b), std::back_inserter(result) );
for ( int x : result )
std::cout << x << "\n";
return 0;
}

Answer 2

You added all the "unique" values from a to the result but you forgot to add all the values from b to the result.
EDIT
Something like that:
for (int i=0; i < sizeof(b)/sizeof(int); ++i, ++r)
{
result[r] = b[i];
}

Answer 3

Your algorithmics is wrong here. For one thing your code runs in o(a*b) where a and b are the list lengths.
Besides, it only produces elements that are in A but not in B.
The way to do this is to sort both lists, then browse through A and B, moving to the smallest current element and adding it to the result if both current elements are different.
This will run in o(log a + log b + a + b).