Running time with while loop

Question

What's the running time of:

foo(n)
if(n==1) return;
int i=1;
while(i<n)
{
    i=i+2
}
foo(n-2)

There are $n/2$ recursive calls to foo but how to add to the calculation the while loop

Possible duplicate of Is there a system behind the magic of algorithm analysis? — Discrete lizard, Jan 20 '18 at 17:50

score 1 · Answer 1 · edited Jan 21 '18 at 01:52

1

The number of the iterations of the while loop is

$$n/2+(n-2)/2 + (n-4)/2 + \dots +1$$

which is $O(n^2)$.

edited Jan 21 '18 at 01:52

D.W.

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answered Jan 20 '18 at 18:07

John P

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What about the $n/2$ calls? so what's the total time? – Error 404 Jan 20 '18 at 18:10
We prefer detailed answers that come with explanation. Where did that expression come from? What's the rationale/justification for it? What approach did you use to get that answer? – D.W. Jan 21 '18 at 01:53

OmG · Accepted Answer · 2018-01-20T19:51:57.470

0

You can write the relation likes the following: $$T(n) = T(n-2) + \frac{n}{2}$$ $$T(n) = \frac{n}{2} + \frac{n-2}{2} + \frac{n-4}{2}+\cdots +\frac{n-(n-2)}{2} = $$ $$\frac{n^2-2(1+2+3+\cdots+(\frac{n}{2}-1))}{2}=$$ $$\frac{n^2-(\frac{n}{2}-1)\times \frac{n}{2}}{2}= \Theta(n^2)$$

edited Jan 20 '18 at 19:51

answered Jan 20 '18 at 19:45

OmG

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Running time with while loop

2 Answers2