@sourav,

As Juvier++ pointed out, it's an **amortized** O(n) algorithm. That's
because each element can be at most popped twice.

Thanks
Yq

On Mon, Jan 3, 2011 at 11:20 AM, sourav <souravs...@gmail.com> wrote:

> @yq Zhang,
>
> To pop if you are going to "pop all from first stack and push into the
> second stack", then does your operation remain "constant time"? Please
> note that we need constant time implementation for the 3 functions
> pop_front, push_rear and get_min(). Goint by your approach, not all of
> them are constant time.
>
> Thanks,
> Sourav
>
> On Jan 3, 9:44 pm, yq Zhang <zhangyunq...@gmail.com> wrote:
> > Push into one stack. When pop first pop all from the first stack and push
> > into the second stack. Then pop from the second stack
> >  On Jan 3, 2011 7:42 AM, "MOHIT ...." <mohit...@gmail.com> wrote:
> >
> > > if only two stack are used but how pop_front is get?
> >
> > > suppose if element comes in order
> >
> > > 12 15 4 3 7 20
> > > then in min queue
> > > 1. 12 (12)
> > > 2. 12 12 (12,15)
> > > 3. 12 12 4 (12,15,4)
> > > 4.12 12 4 3 (12,15,4,3)
> > > 5.12 12 4 3 3 (12,15,4,3,7)
> > > 6.12 12 4 3 3 3 (12,15,4,3,20)
> >
> > > we can get min in constant by pop of stack but how pop front will work
> > using
> > > two stack only in constant time?
> >
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