ok lets think in terms of hashmap of number and frequency
we definitly have to maintain an order array .
traverse original array ,  we check if its present in hashmap ..if not
- 1)add it to hashmap and set frequency to 1 2)add it in order array
if yes then increment frequency by 1.
searching taking O(1) time .

now traverse order array..search for corresponding freq and print

this should take o(n) time...let me know if i faultered anywhere.

@aditi- pls dont use just arrays..its not efficient way.

On Jul 31, 12:51 am, rahul mittal <rahulmitta...@gmail.com> wrote:
> @nivedita:can u do this in o(n) time , i suppose your algorithm takes
> 0(nlogn) time
>
> On Sun, Jul 31, 2011 at 1:14 AM, nivedita arora <vivaciousnived...@gmail.com
>
>
>
>
>
>
>
>
>
> > wrote:
> > @rahul- balanced BST can be maintained.to remove worst case !
> > @sukhmeet- i did not gt your method completely ..u are trying to
> > maintain hashmap or double dimension array of number and count?
> > @aditi- for array we will have overhead of searching if element is
> > present or not ..in O(n) time while it will be O(logn) in BST.
> >  in array we cannot do binary search as it will not be sorted to
> > improve search .
>
> > On Jul 31, 12:31 am, rahul mittal <rahulmitta...@gmail.com> wrote:
> > > maintaining the bst of n element in worst case will take n square time
> > > complexity ..do we have a better solution for worst case
>
> > > On Sun, Jul 31, 2011 at 12:53 AM, sukhmeet singh <sukhmeet2...@gmail.com
> > >wrote:
>
> > > > nivedita: wts the use to maintaining BST.. if the same purpose can be
> > > > fulfilled by an array ..
> > > > but this can be a good method if the range of numbers is pretty large..
> > den
> > > > making a hash count can be difficult..
>
> > > > On Sun, Jul 31, 2011 at 12:43 AM, nivedita arora <
> > > > vivaciousnived...@gmail.com> wrote:
>
> > > >> take a BST whose node has an element of frequency .and another array
> > > >> which will store order of elements.
> > > >> for each array element search BST if node already exists increase the
> > > >> freq count ..other wise  add that element in the order array we took
> > > >> and insert new node in BST.
>
> > > >> now , scan the order array find its corresponding element in BST and
> > > >> its frequency, print it that many times.
>
> > > >> let me know if there is any bttr method
> > > >> thx!
>
> > > >> On Jul 30, 11:58 pm, sukhmeet singh <sukhmeet2...@gmail.com> wrote:
> > > >> > maintain a count array of all elements..
> > > >> > now traverse the array again and the count array .. and build the
> > new
> > > >> array
>
> > > >> > On Sun, Jul 31, 2011 at 12:24 AM, aditi garg <
> > aditi.garg.6...@gmail.com
> > > >> >wrote:
>
> > > >> > > Q1) Given an array with some repeating numbers. Like 12,6,5,12,6
> > > >> > > output: 12,12,6,6,5
>
> > > >> > > 12 shud come before 6 since it is earlier in list. Please provide
> > a
> > > >> > > gud algo fr dis
>
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> > > --
> > > *With Regards *
> > > *RAHUL MITTAL*
> > > *3rd YEAR*
> > > *CSE*
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> *RAHUL MITTAL*
> *3rd YEAR*
> *CSE*
> *MNNIT ALLAHABAD*

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