[algogeeks] Re: Sorting o(n) time

2007-03-19 Thread k3xji

Hi all,
Quoted from Wikipedia:

Pigeonhole sorting, also known as count sort, is a sorting algorithm that 
takes linear time (Θ(n)), which is the best possible performance for a sorting 
algorithm since one must inevitably look at each of the elements being sorted 
at least once, regardless of the sorting algorithm. However, pigeonhole 
sorting is only practical if the objects you are sorting fall within (or can 
be mapped into) a range of possible values that is small compared to the size 
of the list to be sorted.


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[algogeeks] Re: Sorting o(n) time

2007-03-18 Thread BiGYaN

Yeah, after finding the k-th element there is no need for further
partitioning. This is logically true indeed.

But in this case, I guess you have to do it for determining who'll get
the Samosas and who the Gulabjamuns !!.


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[algogeeks] Re: Sorting o(n) time

2007-03-17 Thread Amal
Do  we have to partition after finding the kth element ? Will the Array not
be partitioned when the base condition is met ?

Amal.

On 3/16/07, Rajiv Mathews [EMAIL PROTECTED] wrote:


 On 3/16/07, Karthik Singaram L [EMAIL PROTECTED] wrote:
 
  How would the median algorithm get the top k elements?
 

 I think what he meant was, use the median by parting-in-5 algorithm to
 find the kth largest element. This can be done in O(n).
 Then just partition around this element to get the top k elements, O(n).

 --


 Regards,
 Rajiv Mathews

 


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[algogeeks] Re: Sorting o(n) time

2007-03-16 Thread Rajiv Mathews

On 3/16/07, Karthik Singaram L [EMAIL PROTECTED] wrote:

 How would the median algorithm get the top k elements?


I think what he meant was, use the median by parting-in-5 algorithm to
find the kth largest element. This can be done in O(n).
Then just partition around this element to get the top k elements, O(n).

-- 


Regards,
Rajiv Mathews

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[algogeeks] Re: Sorting o(n) time

2007-03-15 Thread BiGYaN

You could try the quick-sort algo with these further modifications :
every time you partition the list (assuming ascending) you leave out
the entire working for the right hand part iff the size of the left
hand part is = m (m is the top m elements that you need). NB : Here
left hand part indicates the part to the left of the pivot i.e. right
from the beginning of the array.

This on an average case should be able to accomplish the task in O(n)
time.


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[algogeeks] Re: Sorting o(n) time

2007-03-14 Thread chitta koushik
you can sort the students list according to marks in o(n) time by using
RADIX SORT.

On 3/14/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:


 Dean of Computer Science department wants to keep the motivation
 levels of students of data structure course high by giving incentives.
 He has suggested that after the second minor the top  lon students be
 distributed samosas and the next  sqrt(n) students be given gulab
 jamuns. However, he has stipulated that we cannot sort the student
 roll list according to marks (which would také time) but should use
 data structures skills to determine who gets what in O(n) time???


 



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30 years from now it doesn't matter which shoe you wore,which branded jean
you wore..what all matters is WHAT YOU HAVE LEARNED AND HOW YOU HAVE USED
IT.

http://students.iiit.ac.in/~koushik_c

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