Question

It’s from homework, but I’m asking for a general method.

Calculate the following code’s worst case running time.

int sum = 0;
for (int i = 0; i*i < N; i++)
    for (int j = 0; j < i*i; j++)
        sum++;

the answer is N^3/2, could anyone help me through this?

Is there a general way to calculate this?

This is what I thought:

when i = 0, sum++ will be called 0 time
when i = 1, sum++ will be called 1 time
when i = 2, sum++ will be called 4 times
...
when i = i, sum++ will be called i^2 times

so the worst time will be
0 + 1 + 4 + 9 + 16 + ... + i^2

but what next?? I’m lost here…

Answer 1

You want to count how many times the innermost cycle will run.

The outer one will run from i = 0, to i = sqrt(N) (since i*i < N).
For each iteration of the outer one the inner one will run i^2 times.

Thus the total number of times the inner one will run is:

1^2 + 2^2 + 3^2 + ... + sqrt(N)^2

There is a formula:

1^2 + 2^2 + ... + k^2 = k(k+1)(2k+1) / 6 = O(k^3).

In your case k = sqrt(N).

This the total complexity is O(sqrt(N)^3) = O(N^(3/2)).