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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