@sharad in the first step you will have to merge k lists each of n size ..
you will mwrge 2 at a time wouldn't that be O(n*k/2)
so t(n)=n(k/2)+n(k/4)+..
correct me if i'm wrong anyways
On Sun, Jul 4, 2010 at 9:56 AM, Dave wrote:
>
> @Sharad. Haven't you changed the roles of
@dave
u r rite its my fault
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@Sharad. Haven't you changed the roles of n and k? The original poster
had n lists with average length k, meaning a total of n*k elements.
You have k lists with a total of n elements, meaning that each list
has an average of n/k elements.
Dave
On Jul 3, 1:02 pm, sharad kumar wrote:
> 1. First t
