How to calcualte the Big-O complexity of the following algorithm?

Question

I have been trying to calculate the Big-O of the following algorithm and it is coming out to be O(n^5) for me. I don't know what the correct answer is but most of my colleagues are getting O(n^3).

for(i=1;i<=n;i++)
{
    for(j=1 ; j <= i*i ; j++)
    {
        for(k=1 ; k<= n/2 ; k++) 
        {
        x = y + z;
        }
     }
}

What I did was start from the innermost loop. So I calculated that the innermost loop will run n/2 times, then I went to the second nested for loop which will run i^2 times and from the outermost loop will run i times as i varies from 1 to n. This would mean that the second nested for loop will run a total of Sigma(i^2) from i=1 to i=n so a total of n*(n+1)*(2n+1)/6 times. So the total amount that the code would run came out to be in the order of n^5 so I concluded that the order must be O(n^5). Is there something wrong with this approach and the answer that I calculated?

I have just started with DSA and this was my first assignment so apologies for any basic mistakes that I might have made.

Answer 1

The inner loop always has the same number of iterations (n/2), since it is independent of i and j. On its own it has a complexity of O(n).

The two other loops result in a sum of sequence of squares (1 + 4 + 9 + ...) of executions of the inner part.

This sum of squares corresponds to the square pyramidal number, and has an order of O(n³).

The inner loop has an order of O(n), so we get O(n⁴).