Yes, i agree, it will increase the code complexity only. you can just set
the counter to -ve value only.

On Sun, Jul 31, 2011 at 1:38 AM, Kamakshii Aggarwal
<kamakshi...@gmail.com>wrote:

> @neeraj:deleting after printing  will adds to complexity..
>
> On Sun, Jul 31, 2011 at 1:34 AM, Neeraj Gupta 
> <neeraj.gupta...@gmail.com>wrote:
>
>> Create a balance BST.
>> Maintain counter. Whenever You hit duplicate increase the counter while
>> inserting.
>> O(nlogn) for creating it and O(N) space.
>> Now while traverse the array.
>> If you find the element, then print it acco the counter value. After
>> printing delete it.
>> if not found continue traversing.
>> - Show quoted text -
>>
>> On Sun, Jul 31, 2011 at 1:21 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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>
>
>
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> Kamakshi
> kamakshi...@gmail.com
>
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