Question

This is my code for a simple selection sort.usually the complexity (time) for a sort is number of iterations it has taken for sorting O(n^2) in case of selection sort
When I dry ran this code against sample string of 98765, it gave me 25 iterations.
Just to cross check with my dry ran output i put 2 vbl- noi and noj in my code.

Q: will the number of total iterations be = noi*noj or noi+noj;

int index = 0; int noi = 0, noj = 0;

for (j = 0; j < 5; j++)
{
    noj++;
    index = j;
    for (i = j; i < 5; i++)
    {
        if (a[index] > a[i])
        {
            a[index] = a[index] + a[i];
            a[i] = a[index] - a[i];
            a[index] = a[index] - a[i];
            noi++;
        }

    }

}

Answer 1

number of iterations is always 15 (5+4+3+2+1) because in your loops there are j<5 and i<5. So your code complexity is O(n^0) because in your case n is 5