Karumanchi

Question

 

Comments

n(logn)

Answer 1

It should be $O(n)$.

It can be written as:
 

while(n > 1){
    for(i = 1; i <= n; i++)
        temp = temp + 1;
    n = n / 2;
}

It should be $O(n)$. What I forgot was that the inner value of $n$ will also change.

So the first time, it will run $n$ times, then $n/2$ times and so on, thus making it $O(n).$
 

Comments

n+n/2+n/4+n/8+…n+n/2+n/4+n/8+… up to log⁡n terms

 = 2(n-1)

 kumar.dilip how you right directly 2(n-1) i am not able to understand

it is a GP i can not solve it ,eloborate it 

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you can cross check it. By putting the value of n.

or follow this link 

https://math.stackexchange.com/questions/244817/solve-n-n-2-n-4-n-8-dots-up-to-log-n-terms

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thanks kumar.dilip

Answer 2

The answer should be O(n) as the inside value of n will change as well.