its not exponential
time to find out m = m
time to create list 3 = m
time to create list 4 = n-m
time to come up with proper added list (list 3 modification) = m
time to merge list 3 and list 4 = n-m

total time = 2n+m

except step 1 all are reversals with approximately same constant and
constant for step 1 is smaller
so one can say

O(2n+m)

On Wed, Jan 27, 2010 at 5:26 PM, Anurag Bhatia <abhati...@gmail.com> wrote:

> If that is the representation, then the lists have to be reversed.
> Otherwise the time goes up exponentially.
>
> On Wed, Jan 27, 2010 at 5:19 PM, Algoose Chase <harishp...@gmail.com>
> wrote:
> > Condition:
> > Can we do it keeping the original lists intact ? ie without reversing it.
> > I mean , No recursion & no Reversing ... is it possible ?
> >
> > @kumar :
> > 15234 is represented as  1->5->2->3->4
> >
> > On Wed, Jan 27, 2010 at 4:09 PM, saurabh gupta <sgup...@gmail.com>
> wrote:
> >>
> >> perhaps you mean,
> >> reverse each link list O(n+m).
> >> then sum each node with carryover maintained.
> >>
> >> On Wed, Jan 27, 2010 at 11:07 AM, Anurag Bhatia <abhati...@gmail.com>
> >> wrote:
> >>>
> >>> Let us take an example -
> >>>
> >>> Num 1 = 123456
> >>> Num 2= 1234
> >>> Link-1->Link-2->Link-3->Link-4->Link5->Link6
> >>> Link-1->Link-2->Link-3->Link-4
> >>>
> >>> Add nodes into linkedlist 1 till either one of the list is not null.
> >>> Make sure you process the carry in each iteration.
> >>>
> >>>
> >>> --AB
> >>>
> >>>
> >>> On Tue, Jan 26, 2010 at 9:47 PM, Algoose Chase <harishp...@gmail.com>
> >>> wrote:
> >>> > conditions:
> >>> > NO extra memory (@ stack or Heap) at all. No recursion.
> >>> >
> >>> > Any body has got any hint about how to get this done ?
> >>> >
> >>> >
> >>> >
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