Question

I have problem in determining time complexities of algorithms.

for(int i=0;i <n i++){}   O(n)

for(int i= 0 ;i<n ;i++){    O(n^2)
  for(int j=0;j<n;j++){ 

  }
}

Now for following code whats the complexity

for(i =0; i<n ; i++) {}
for (j=0;j<n ;j++ ) {} 

is it O(2n) as it invloves 2 seperate loops?

what if i start j =5 to n?

Answer 1

There is no O(2n), it’s just O(n). In other words, it scales at the same rate as n increases.

If it was a nested loop, it would be O(n2) but the presence of your {} empty blocks means it isn’t nested.

And it makes no difference whether you start at one or five, it still scales with n, just with a slightly negative constant addition. Hence still O(n).

The complexities O(n), O(cn) and O(n+c) (where c is a constant) are all equivalent. In addition, you also generally only use the term with the highest effect.

So you won’t usually see O(7n3 + 3n2 + 12n + 2), that will be simplified to O(n3).