@dheeraj - for count sort we need to know the range the numbers are in.
Otherwise how will u initialize array keeping count?

On Sun, Sep 4, 2011 at 12:03 AM, Dheeraj Sharma <[email protected]
> wrote:

> count sort in reverse order would help i guess :)
>
>
> On Sat, Sep 3, 2011 at 11:49 PM, mohit verma <[email protected]>wrote:
>
>> Ohh my bad. the complexity should be
>> O(nlogn) + O(mlogm) +O(m) = O(tlogt) where t=max(m,n)
>>
>>
>> On Sat, Sep 3, 2011 at 11:46 PM, mohit verma <[email protected]>wrote:
>>
>>> create a binary search tree by scanning the whole array and if the value
>>> already exists increase count field in that node O(nlogn). Now traverse the
>>> tree in any order by creating another tree with kery - count. O(nlogn).
>>> Doing reverse inorder traversal print value field of each node count number
>>> of times O(n).
>>>
>>> overall complexity - O(nlogn)+O(nlogn)+O(n) = O(nlogn).
>>>
>>>
>>> On Sat, Sep 3, 2011 at 11:16 PM, Siddhartha Banerjee <
>>> [email protected]> wrote:
>>>
>>>> perhaps we can make it O(mlogm), m= no of distinct elements... just
>>>> create a hash table and store the count of the number of time elements
>>>> occur... O(n), now sort the m elements and proceed as above...
>>>> it is obviously not possible to do it faster than O(mlogm), where m = no
>>>> of distinct elements...
>>>> so order = O(n+mlogm)=O(mlogm)(???, assuming m!<<n)
>>>>
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>>>
>>>
>>>
>>> --
>>> ........................
>>> *MOHIT VERMA*
>>>
>>>
>>
>>
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>> *MOHIT VERMA*
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>
>
> --
> *Dheeraj Sharma*
> Comp Engg.
> NIT Kurukshetra
> +91 8950264227
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