@Richa
Okey..
For this scenario,
Extracting a node => output the root node, decrement its frequency
count and remove the root node only if frequency count becomes 0.
Heapify.
Adding a node => locate a node which corresponds to same character as
the inserted char in O(N) time, if it exists and increment its
frequency count. Else create a new node with frequency count 1 and
character = <new_char_input_by_user>  and add i to the heap.

Given the above description above I think the input cases pointed out
by you can be worked out.
_dufus



On Aug 17, 12:15 am, richa gupta <richa.cs...@gmail.com> wrote:
> @dufus :
>
> suppose the words occurs like ( at the last of the document or u can say
> that in the window of size N , the scenario is )
> l, b,b,b, c , d,d, e a
> so ur max heap will make it
>            b(3)
>   d(2)                    c(1)
> e(1)   a (1)             l(1)
>
> now explain me what happens when b or d or some new word "m"
>
> 2009/8/16 Miroslav Balaz <gpsla...@googlemail.com>
>
>
>
> > I think the teoreticaly easiest solution is to have two balanced binary
> > search trees interlinked. on for finding by string, second for keeping count
> > second can keep subtree sizes, and every time you check the position and if
> > it changes then you can update list, you do not have to make more
> > computation than is needed to redraw screen and to maintain balanced tree.
>
> > 2009/8/14 richa gupta <richa.cs...@gmail.com>
>
> >> You have to develop a piece of code that can be attached to a program
> >> like Microsoft Word, which would list the last "N" words of the
> >> document in descending order of their occurence, and update the list
> >> as the user types. What code would you write? Using pseudo code is
> >> fine, just list the basic things you would consider
>
> >> --
> >> Richa Gupta
> >> (IT-BHU,India)
>
> --
> Richa Gupta
> (IT-BHU,India)

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