Question

For homework, I was given the following 8 code fragments to analyze and give a Big-Oh notation for the running time. Can anybody please tell me if I’m on the right track?

//Fragment 1 for(int i = 0; i < n; i++)     sum++; 

I’m thinking O(N) for fragment 1

//Fragment 2 for(int i = 0; i < n; i+=2)     sum++; 

O(N) for fragment 2 as well

//Fragment 3 for(int i = 0; i < n; i++)     for( int j = 0; j < n; j++)         sum++; 

O(N^2) for fragment 3

//Fragment 4 for(int i = 0; i < n; i+=2)     sum++; for(int j = 0; j < n; j++)     sum++; 

O(N) for fragment 4

//Fragment 5 for(int i = 0; i < n; i++)     for( int j = 0; j < n * n; j++)         sum++; 

O(N^2) for fragment 5 but the n * n is throwing me off a bit so I’m not quite sure

//Fragment 6 for(int i = 0; i < n; i++)     for( int j = 0; j < i; j++)         sum++; 

O(N^2) for fragment 6 as well

//Fragment 7 for(int i = 0; i < n; i++)     for( int j = 0; j < n * n; j++)         for(int k = 0; k < j; k++)             sum++; 

O(N^3) for fragment 7 but once again the n * n is throwing me off

//Fragment 8 for(int i = 1; i < n; i = i * 2)     sum++; 

O(N) for fragment 8

Answer 1

I think fragment 5 is O(n^3), and similarly fragment 7 is O(n^5)*. It also looks like O(log(n)) for fragment 8.

For the n * n problems, you have to execute the body of the loop n * n times, so it would be O(n^2), then you compound that with the order of the other code. Fragment 8 actually doubles the counter instead of incrementing it, so the larger the problem, the less additional work you have to do, so it’s O(log(n))

*edit: Fragment 7 is O(n^5), not O(n^4) as I previously thought. This is because both j and k go from 1 to n * n. Sorry I didn’t catch this earlier.