Question

for(i = 1; i < n*n; i++){
    for(j = 1; j < i*i; j++){
        if(j % i == 0){
            for(k=0; k < j; k++){
                count++;
            }
        }
    }
}

My attempt at a solution:

j iterates up to i*i = n^4. For the ‘k’ loop, we have the sum of k from 1 to n^4 which is n^4(n^4-1)/2. So the runtime is O(n^8). This strikes me as too high, but I don’t see an error.

Answer 1

The outer loop executes n² times. The next loop executes a total of ∑_i=1ⁿ² i², which is O(n⁶).

The innermost loop only runs when j is a multiple of i, which happens i times for each value of i. The innermost loop executes j times for each such value of j: i, 2i, 3i, and so forth until i*i. Thus, the innermost loop executes ∑_j=1ⁱ ij times, which is O(i³), for each i.

Therefore, the total running time is ∑_i=1ⁿ² O(i³) + O(n⁶), which is O(n⁸) since ∑_i=1ⁿ² O(i³) = O(n⁸).

(Note that I’m assuming the second loop increments j++, not i++. The answer is rather different if it’s i++).