Given an array return any index on which its dominator occurs

Dominator is a value which occurs on more than half positions in an array. For instance in the array: [3, 4, 3, 2, 3, -1, 3, 3] the value 3 occurs in 5 out of 8 positions. 5/8 > 0.5, so 3 is a dominator: in this array the dominator occurs on indexes : 0, 2, 4, 6 , 7. How to write a function that given an array returns any index on which its dominator occurs. If there is no dominator, the function should return -1.

Tried to solve it using C as:

int arrdominator(int *A,int N)
{
    int i,j,idx=0,val=0,cnt=1;

    //I thought I would sort the array and then cound the dominating value instances,
    qsort(A,N,sizeof(A[0]),sort_fn_ascend);

    for(i=0;i<(N-1);i++)
    {
        for(j=i+1;j<N;j++)
        {
            if(A[i] == A[j])
            {
                cnt++;

                i++;
                break;
            }
        }
    }
}

But i could not get to a solution of finding the dominator value in the array and then return its index.

Thanks to @DocBrown's pointers got my initial code refined to get it working alright:

int arrdominator(int *A,int N)
{
    int i,j,idx=0,cnt=1,valmaxseq,dominantval=0;
    int *newA = (int*)malloc(sizeof(int)*N);
    for(i=0;i<N;i++)
        newA[i] = A[i];

    qsort(newA,N,sizeof(A[0]),sort_fn_ascend);


    for(j=0;j<(N-1);j++)
    {
        if(newA[j] == newA[j+1])
        {
            cnt++;              
        }
        else
        {               
            if(cnt > (N/2))
            {                   
                valmaxseq = newA[j];
                dominantval = 1;
            }

            cnt = 1;                
        }
    }

        //find index of dominant value if there is one
        if(dominantval == 1)
        for(i=0;i<N;i++)
        {
            if(A[i] == valmaxseq)
                return i;
        }

        return -1;
}