What is the Big O analysis of this algorithm?（Stackoverflow）

Question：

I'm working on a data structures course and I'm not sure how to proceed w/ this Big O analysis:

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

My initial idea is that this is O(n^3) after reduction, because the innermost loop will only run when 

j

/

i

 has
no remainder and the multiplication rule is inapplicable. Is my reasoning correct here?

Answer：

Let's ignore the outer loop for a second here, and let's analyze it in terms of 

i

.

The mid loop runs 

i^2

 times,
and is invoking the inner loop whenever 

j%i
== 0

, that means you run it on 

i,
2i, 3i, ...,i^2

, and at each time you run until the relevant 

j

,
this means that the inner loop summation of running time is:

i + 2i + 3i + ... + (i-1)*i  = i(1 + 2 + ... + i-1) = i* [i*(i-1)/2]

The last equality comes from sum of arithmetic progression. 

The above is in 

O(i^3)

.

repeat this to the outer loop which runs from 

1

 to 

n

 and
you will get running time of 

O(n^4)

,
since you actually have:

C*1^3 + C*2^3 + ... + C*(n-1)^3 = C*(1^3 + 2^3 + ... + (n-1)^3) =
= C/4 * (n^4 - 2n^3 + n^2)

The last equation comes from sum of cubes 

And the above is in 

O(n^4)

,
which is your complexity.