@Apoorve: I don't believe that you can determine the frequency counts
in-place in O(n).

Dave

On May 9, 8:42 am, Apoorve Mohan <apoorvemo...@gmail.com> wrote:
> @Dave: Can there be an in-place O(n) solution to this???
>
>
>
>
>
> On Mon, May 9, 2011 at 7:01 PM, Dave <dave_and_da...@juno.com> wrote:
> > @Ashish: By compress, I meant to transfer the active entries in the
> > hash table into contiguous elements of an array. Since the hash table
> > itself is probably stored in an array, it just means to go through the
> > hash table array and move the active entries to the front of the
> > array.
>
> > Dave
>
> > On May 9, 1:14 am, Ashish Goel <ashg...@gmail.com> wrote:
> > > Dave,
> > > w.r.t statement, After all integers are processed, compress out the
> > unused
> > > hash table
> > > entries and find the Kth largest element,
>
> > > I could not understand the compress concept...are you saying something on
> > > counting sort philosophy?
> > > say the list is
>
> > > 5
> > > 4
> > > 4
> > > 3
> > > 3
> > > 3
> > > 2
> > > 2
> > > 2
> > > 2
> > > 1
> > > 1
> > > 1
> > > 1
> > > 1
>
> > > so the hash map will have
>
> > > 5,1 <5 is 1 time>
> > > 4,2,<4 is two time>
> > > 3,3,...
> > > 2,4,...
> > > 1,5...
>
> > > so if the question is to find say 3rd largest element based on frequency,
> > it
> > > would be 4
> > > This essentially means that we probably need dynamic order statistics
> > where
> > > the node contains the starting rank for a particular entry in the RBTree.
> > > Each RBTree node contains the number, its frequency and rank. then this
> > > becomes O(n) +O(lgn) i.e. O(n)
>
> > > Best Regards
> > > Ashish Goel
> > > "Think positive and find fuel in failure"
> > > +919985813081
> > > +919966006652
>
> > > On Sat, May 7, 2011 at 11:03 PM, Dave <dave_and_da...@juno.com> wrote:
> > > > @Gaurav: As I understand your solution, you are finding the K largest
> > > > integers based on value, rather than the K with the highest frequency
> > > > of occurrence.
>
> > > > @Amit: Hash the integers into a hash table that includes a field for
> > > > the frequency count. When a new entry is made, set the frequency count
> > > > to 1. When an integer already is in the table, increment its count.
> > > > After all integers are processed, compress out the unused hash table
> > > > entries and find the Kth largest element. The overall algorithm can be
> > > > done in O(n).
>
> > > > Dave
>
> > > > On May 7, 12:06 pm, Gaurav Aggarwal <0007gau...@gmail.com> wrote:
> > > > > It can be done without sorting. Take the first element as pivot
> > > > > element. Calculate its position using Partition algorithm.
> > > > > If its new index is K, then on left side are top K  integers. If
> > index>K,
> > > > > apply algo on left side Else apply algo on right side.
>
> > > > > Best Case: O(1)
> > > > > Worst Case: O(n^2)
>
> > > > > But there are very less chances of worst case. It is when the list is
> > > > > already sorted.
>
> > > > > On Thu, May 5, 2011 at 1:06 AM, amit <amitjaspal...@gmail.com>
> > wrote:
> > > > > > Hi ,
>
> > > > > > We are give a list of integers now our task is to find the top K
> > > > > > integers in the list based on frequency.
>
> > > > > > Apart from sorting the data can there be any other algorithm?
>
> > > > > > --
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> > > > > --
> > > > > Gaurav Aggarwal
> > > > > SCJP- Hide quoted text -
>
> > > > > - Show quoted text -
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> regards
>
> Apoorve Mohan- Hide quoted text -
>
> - Show quoted text -

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