Its nlogn dude

On Wed, Oct 31, 2012 at 8:03 PM, jai gupta <[email protected]> wrote:

> is O(long n!)
> n! is O(n^n) by sterling approximation
> hence it is O(log n^n) = O(nlogn)
>
>
> On Sun, Oct 28, 2012 at 11:14 PM, Pralay Biswas <
> [email protected]> wrote:
>
>> @ Siddharth :
>> Well, here is how you may understand it:
>> 1) There is an outer loop that runs n times. (k)
>> 2) Then there is an inner loop where j is initially set to current k, but
>> it halves itself in every iteration
>>           -- So for example, if k is 32, and j halves every time, then
>> the inner loop runs for 5 times (32->16->8->4->2 etc)
>>           -- This is a log base 2 order depreciation in for all k
>> 3) So for every k, the inner loop runs for log k times. k itself varies
>> from 1 to n, hence O(nlogn)
>>
>> Hope this helps.
>>
>> #Pralay
>> MS-CS, UC Irvine
>>
>> On Sun, Oct 28, 2012 at 10:38 AM, bharat b 
>> <[email protected]>wrote:
>>
>>> while loop : logj base 2 ..
>>> ==> log1 + log2 + ... logn = log(n!) [since logab = loga + logb]
>>> ==> O(log(n^n)) = O(nlogn)
>>>
>>>
>>> On Sun, Oct 28, 2012 at 3:56 AM, Siddharth Malhotra <
>>> [email protected]> wrote:
>>>
>>>> Can somebody explain how it is O(n log n).
>>>> What is the significance of while loop in the above code?
>>>>
>>>> I understand that the for loop implies O(n),does the log n in the O(n
>>>> log n) comes from the while loop?
>>>> What if there where two while loops in the for loop separately?
>>>>
>>>> On Sat, Oct 27, 2012 at 3:26 AM, Pralay Biswas <
>>>> [email protected]> wrote:
>>>>
>>>>> Yes!
>>>>>
>>>>>
>>>>> On Fri, Oct 26, 2012 at 2:55 PM, 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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>
>
>
> --
>
> kriateive
>
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