randomly reorder elements in an array without duplication - c++

Question

I have a fixed array of sized nine and I am trying to reorder it randomly without duplication.
this is the following code:
class numbers{
int randomIndexCount;
public:
void randomArray( int numArray[],int size){
randomIndexCount = 0;
for (int i = 0; i < size; i++)
{
int RandomIndex = rand() % size;
randomIndexCount++;
numArray[i] = numArray[RandomIndex];
cout << numArray[i] <<endl;
}
}
int main(){
srand(time(0));
int numArray[9]= {1,2,3,4,5,6,0,0,0};
numbers n;
n.randomArray(numArray,9);
return 0;
}
So far I was able to reorder the array randomly with the given elements however I am unsure how to get rid of duplication. the output should be {1,2,3,4,5,6,0,0,0} but in a random order.
I am unable to use the shuffle function and can only use rand.
I am not sure how to remove duplicate entries
this is what I had in mind
1) with the given index check if that value already exist and if it does then skip this line " numArray[i] = numArray[RandomIndex];". however this approach would not be efficient as im sure this would be too time consuming.
is there a way to remove duplicate values so my output is something like:
{0,1,0,6,2,0,5,3,4}

Answer 1

You should swap elements in this line numArray[i] = numArray[RandomIndex];, not assigning. This will duplicate data! Here's the swap:
int v = numArray[i];
numArray[i] = numArray[RandomIndex];
numArray[RandomIndex] = v;

Answer 2

You are duplicating the elements with the assignement inside the for loop
numArray[i] = numArray[RandomIndex]
Instead asign the element to the position of the array, you need to swap those elements as follow:
class numbers
{
int randomIndexCount;
public:
void randomArray (int numArray[], int size)
{
randomIndexCount = 0;
// Use srand with a time seed value in order to
// have different results in each run of the programm
srand (time (NULL));
for (int i = 0; i < size - 1; i++)
{
int swap = numArray[i];
//take a random index from 0 to i
int j = rand () % (size);
numArray[i] = numArray[j];
numArray[j] = swap;
cout << numArray[i] << endl;
}
}
};
int main ()
{
int numArray[9] = { 1, 2, 3, 4, 5, 6, 0, 0, 0 };
numbers n;
n.randomArray (numArray, 9);
return 0;
}
This will be the output that includes all then numbers in the array:
5
2
0
6
3
0
1
0

Answer 3

This loop scans positions from the end of the array towards its beginning and randomly selects a new item to be put at the current position. Items are chosen from positions not scanned yet.
This way every place is exactly once chosen to be filled with some item, and each item is exactly once placed in its final position (although, before this happens, it may be several times swapped out of places chosen for other elements).
It also guarantees no item disappears (gets overwritten) and no duplicates appear (no item is copied inadvertently) - if you have duplicates in input data, the same duplicates remain in the output (although permuted); if there are no duplicates, there will be no duplicates.
Additionally, if the rand() function has no bias, every item has the same chance to end at any chosen position, hence each possible permutation is equally probable as an output.
for (int i = size; i > 1; -- i)
{
int swapIndex = rand() % i;
int swap = numArray[swapIndex];
numArray[swapIndex] = numArray[i-1];
numArray[i-1] = swap;
}