@souravsain :Your approach works really well..

Here is the Implementation:
http://codepad.org/ricAcQtu



On Sun, Jun 6, 2010 at 11:40 AM, souravsain <souravs...@gmail.com> wrote:

> @divya:go through the elements and keep inserting them in a BST. While
> inserting if elements already exists in BST, increase its frequency
> (Node of BST has element a nd frequency). Also if elemengs is newly
> inserted then also place it in a seperate array. So this array (say
> Array M) will become something like 12,5,6. This array will give order
> of first occurance of numbers. This whole process will take nlogn (BST
> creation assuming worst case that all elements are uinque in the input
> array).
>
> Once done, scan through each element in array M, find its
> corrosponding element in BST (logn) which will give the frequency.
> Fill those many indexes in input array with array M[i]. If all
> elements are uinque, this will also take nlogn. Else if input array
> has m distince elements, which will require us to look into BST for m
> times.
>
> hence entire process has time compelxity: O(nlogn + nlogn)= O(2nlogn)
> Space complexity: O(2n) [1 for BST and 1 for array M, worst case when
> all elements are unique in inpur array).
>
> Let me know your comments if any or any better appraoch as this once
> may have improvements.
>
> On Jun 6, 7:47 pm, divya jain <sweetdivya....@gmail.com> wrote:
> > output willl be 12 12 5 6 6
> >
> > On 6 June 2010 18:27, souravsain <souravs...@gmail.com> wrote:
> >
> >
> >
> > > @divya: Does your problem require the output to be sorted also? What
> > > will be the output required if inout is 12,5,6,12,6? Will it be
> > > 12,12,6,6,5 or 12,12,5,6,6,?
> >
> > > Sain
> >
> > > On Jun 6, 12:01 am, divya <sweetdivya....@gmail.com> wrote:
> > > > 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. So cant use a
> > > > dictionary
> >
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