Question

So, I was searching for a good solution for my problem.

I need to generate(print) all the combination of a list of integers, for example:
if the array contain integers from 0 to n-1, where n = 5:

int array[] = {0,1,2,3,4};

the order of integers in the combination are NOT important, meaning {1,1,3}, {1,3,1} and {3,1,1} are actually the same combination because they all contain one 3 and two ones.

so for the above array, all combination of length 3:

0,0,0 -> the 1st combination
0,0,1
0,0,2
0,0,3
0,0,4
0,1,1 -> this combination is 0,1,1, not 0,1,0 because we already have 0,0,1. 
0,1,2
0,1,3
0,1,4
0,2,2 -> this combination is 0,2,2, not 0,2,0 because we already have 0,0,2. 
0,2,3
.
.
0,4,4
1,1,1 -> this combination is 1,1,1, not 1,0,0 because we already have 0,0,1. 
1,1,2
1,1,3
1,1,4
1,2,2 -> this combination is 1,2,2, not 1,2,0 because we already have 0,1,2.
.
.
4,4,4 -> Last combination

For Now I Wrote Code for doing this, but my problem is:
if the numbers in the array are not integer from 0 to n-1, lets say if the array was like this

int array[] = {1,3,6,7};

my code doesn’t work on this case, any algorithm or code for solving this problem,,

Here is my code :

unsigned int next_combination(unsigned int *ar, int n, unsigned int k){
    unsigned int finished = 0;
    unsigned int changed = 0;
    unsigned int i;

    for (i = k - 1; !finished && !changed; i--) {
        if (ar[i] < n - 1) {
            /* Increment this element */
            ar[i]++;
            if (i < k - 1) {
                /* Make the elements after it the same */
                unsigned int j;
                for (j = i + 1; j < k; j++) {
                    ar[j] = ar[j - 1];
                }
            }
            changed = 1;
        }
        finished = i == 0;
    }
    if (!changed) {
        /* Reset to first combination */
        for (i = 0; i < k; i++){
            ar[i] = 0;
        }
    }
    return changed;
}

And this is the main:

int main(){
    unsigned int numbers[] = {0, 0, 0, 0, 0};
    const unsigned int k = 3;
    unsigned int n = 5;

    do{
        for(int i=0 ; i<k ; ++i)
            cout << numbers[i] << " ";
        cout << endl;
    }while (next_combination(numbers, n, k));

    return 0;
}

Answer 1

If you have working code to generate all combinations of numbers from 0 to n-1, then this is very simple. You have your array of numbers:

int array[] = {1,3,6,7};

Now, take n = 4, because there are 4 items in the array. Generate all combinations from 0 to 3, and use those as indices into your array. You now have all combinations of your array values by using all combinations of indices into that array.