@nivedita :hashing will not work if the range of nos is high

On Sun, Jul 31, 2011 at 1:40 AM, nivedita arora <vivaciousnived...@gmail.com
> wrote:

> 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*
> > > > *MNNIT ALLAHABAD*
> >
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> > --
> > *With Regards *
> > *RAHUL MITTAL*
> > *3rd YEAR*
> > *CSE*
> > *MNNIT ALLAHABAD*
>
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-- 
Regards,
Kamakshi
kamakshi...@gmail.com

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