count sort in reverse order would help i guess :)

On Sat, Sep 3, 2011 at 11:49 PM, mohit verma <mohit89m...@gmail.com> 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 <mohit89m...@gmail.com>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 <
>> thefourrup...@gmail.com> 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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-- 
*Dheeraj Sharma*
Comp Engg.
NIT Kurukshetra
+91 8950264227

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