u people absolutely correct that log operation take much time , i just
given an approach
On Sat, Dec 10, 2011 at 7:25 AM, Dave wrote:
> @Gaurav: Even though this is O(log n), it is bound to be slow, since
> it would use division and modulus are among the slowest operations on
> modern computers,
Originaly problem rules out the use of log.Moreover log (or any floating
point operations) take lot of hardware time as they are emulated on the
floating point environment on most machines.Thirdly precision problem for
higher values of n may cause your solution to give wron g answers,,,
On Wed, De
what about this log(base 3 n) is of integral type then n is a power of 3
On Mon, Dec 5, 2011 at 11:36 PM, Dave wrote:
> @Carl: You can generate the constants the first time the procedure is
> called. There is no need to do them every time. So the first call
> would be O(wordsize) but subsequ
Sorry, I was being a bit vague. I meant what would the time complexity be
then?
As I understand your Constant time is Dependant on the constant pre
computation you do, which is no longer the case when you generalise
On 5 December 2011 16:14, Dave wrote:
> @Carl: Of course. For any given word siz
Ah I see, in which case could you not generalise your solution for all
integers?
By taking into account the size of words on the computer for example?
On 5 December 2011 15:09, Dave wrote:
> @Carl: Yes, as coded, my algorithm is for 32-bit integers. But the
> original poster asked for a soluti
@Dave Yours only works for a certain subset of all possible powers or 3
doesn't it? So WgpShashank's would be more general?
On 5 December 2011 14:30, Dave wrote:
> @WgpShashank: Yours is an O(log n) solution. Mine is O(1).
>
> Dave
>
> On Dec 5, 6:21 am, WgpShashank wrote:
> > @SAMMM have a lo
+1 nitin
there must be n-1 dividers
On Wed, Aug 17, 2011 at 4:15 PM, Puneet Goyal wrote:
> I think it should be 2^n -1
>
> Explanation
> We can visualize it as n balls are placed and we have to place some
> dividers (max=n) in betweek to divide them into groups.
>
> If we choose no divider its nC
@Puneet, you are right but we can have only n-1 dividers.
On Wed, Aug 17, 2011 at 4:15 PM, Puneet Goyal wrote:
> I think it should be 2^n -1
>
> Explanation
> We can visualize it as n balls are placed and we have to place some
> dividers (max=n) in betweek to divide them into groups.
>
> If we ch
I think it should be 2^n -1
Explanation
We can visualize it as n balls are placed and we have to place some dividers
(max=n) in betweek to divide them into groups.
If we choose no divider its nC0 , but we dont have to include it
With 1 divider its nC1
and so on..
So the total no. of ways will be
+1 to nitin
On Wed, Aug 17, 2011 at 2:48 PM, Vijay Kansal wrote:
> my bad 2^(n-1)...
>
> On Aug 17, 2:17 pm, Vijay Kansal wrote:
> > @nitin it must be 2^n i think
> >
> > On Aug 17, 3:48 am, Bharat Kul Ratan wrote:
> >
> >
> >
> >
> >
> >
> >
> > > It might be useful:
> http://www.artofproblems
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