yeah..we it works for +ve  number..within some specified range..it was
alternate to O(nlogn) solution..if range is known

On Sun, Sep 4, 2011 at 12:07 AM, mohit verma <[email protected]> wrote:

> @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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>> --
>> *Dheeraj Sharma*
>> Comp Engg.
>> NIT Kurukshetra
>> +91 8950264227
>>
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-- 
*Dheeraj Sharma*
Comp Engg.
NIT Kurukshetra
+91 8950264227

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