Yes, its complexity is O(nlogn).

On Sat, Oct 27, 2012 at 2:35 PM, CHIRANJEEV KUMAR
<[email protected]>wrote:

> O(log( n! ));
>
>
> On Sat, Oct 27, 2012 at 6:03 AM, payal gupta <[email protected]> wrote:
>
>> That 's correct.
>>
>>
>> On Sat, Oct 27, 2012 at 3:25 AM, rahul sharma <[email protected]>wrote:
>>
>>> for k=1 to n
>>> {
>>> j=k;
>>> while(j>0)
>>> j=j/2;
>>> }
>>> the complexity is big o is o(nlogn)
>>> am i ryt????
>>>
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-- 
Anil Kumar Sahu
M. Tech.(Information Security)
MNNIT Allahabad
Ph. No:+91 8123065616, 8303325388

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