Vector to vector without repeating elements

[nenko182]

Basically, I have a vector with 10 elements, and I need to create another vector with 6 non-repeating elements, taken randomly from the first vector. Kind of like a 6-combination. The thing is, though, that I cannot successfully exclude the repeating elements and I always get repeated elements in my 6-element vector.

I tried several things: nested loop with an iterator to check content of the vector, I tried removing the element, which I just inserted into the second element - I get crashes or repeated elements in the end.

What is the best approach? Something I haven't tried yet is using binary search.

Here are some attempts I did, which did not work. I do understand why they do not work, but not well enough to improve these solutions and make it work.

//One try - vec3 is a vector if integers
    vector<int>::iterator iter;
    srand(time(NULL));
    vector<int> sixComb;
    for(int i=0; i<6; i++)
    {
        int k=1+rand()%10;
        int numb = vec3[k];
        if (sixComb.size() == 0)
        {
            sixComb.push_back(numb);
        }
        for (iter=sixComb.begin(); iter != sixComb.end(); ++iter)
        {
            if (numb == *iter) break;
            if (numb != *iter) sixComb.push_back(numb);
        }
    }
//Second try
    vector<int>::iterator iter;
    srand(time(NULL));
    vector<int> sixComb;
    for(int i=0; i<6; i++)
    {
        int k=1+rand()%10;
        int numb = vec3[k];
        do
        {
            sixComb.push_back(numb);
        }
        while (numb!=sixComb[i]);

    }
//Three - I tried deleting the element I have just inserted
    for(int i=0; i<6; i++)
    {
        int k=1+rand()%10;
        int numb = vec3[k];
        sixComb.push_back(numb);
        vec3.erase (vec3.begin()+k);

    }

[ne555]

bool is_member(element, container){
   return find( container.begin(), container.end(), element ) not_eq container.end();
}

repeat(6){
   while( is_member( vec[K], sixComb )
      K = give_me_other();
   sixComb.push_back(vec[K]);
}

[jlb]

What happens if there are not 6 non-repeating elements in your original vector?

Can you sort the the original vector, or must the order be maintained?

By the way, I myself would remove the call to srand() until you have the program working for one set of numbers.

[nenko182]

The elements in my original vector are fixed and there are NO repeated elements in there.
The order does not matter, the resulting 6-element vector would eventually be sorted.

[Duthomhas]

Copy all the elements of the original vector into a new vector. Randomize the new vector. Truncate it to N elements. Sort.

[nenko182]

SOLVED

@Duoas how would that assure me that there will be no repeated elements in the new vector?

Edit: I think I understand - because of the fact that I just use the original vector's content once, without copying repeatedly differeny, random elements N times.

[TarikNeaj]

Each time, before you copy over one element, loop through your vector and make sure it doesnt exist there already.

[Duthomhas]

Sorry, I assumed that your source vector has no duplicates.