Hi SAMM,

The above code is not clear enough to understand  . Sorry for that.
My idea was , like for above example : map will contain frequency of all
distinct elements.

so  freq[1] = 6
     freq[3]= 1
    freq[5]=1
   freq[6]=1

Now for n=9 and k=3
 P1={1,3,5};
and now after this partition frequency of each element gets reduced by
1.Now you can eliminate elements having 0 frequency or just skip them.

In second run
P2={1,6,1}

and P3={1,1,1}.


On Mon, Oct 31, 2011 at 4:20 AM, SAMM <[email protected]> wrote:

> Ur algo will not work for this case :-
>
> 1 1 1 1 1 1 3 5 6   ---- For the array .. And for K=3
>
> Ur algo will give  (1 1 1) (1 1 1 ) (3 5 6)
>
> On 10/30/11, mohit verma <[email protected]> wrote:
> > sort the array : O(nlogn)
> >
> > keep an array/map containing frequency of each element in sorted array :
> > O(n)
> >
> > v[n/k][k] - 2-D array of ints  containing final partitions.
> >
> > for i=1 to n/k
> >    {
> >      int count=0;
> >      for(j=0;j<n && count < k;j++)
> >        { if(  freq(a[i])==0) continue; //say array is used
> >            v[i][count]=a[i];
> >            freq(a[i])--; //just an idea , not actual implementation
> >             count++;
> >        }
> > }
> >
> > you can improve internal for loop by using map : if  freq[a[i]] becomes 0
> > delete the node from array.
> > On Sun, Oct 30, 2011 at 10:35 PM, SAMM <[email protected]> wrote:
> >
> >> No there is no such condition ...just hav to make sure all the
> >> partitions are unique .
> >> The partitions can hav some elements (< K) in common but not the
> >> entire elements in a partition (Should be UNIQUE) .
> >>
> >> On 10/30/11, sunny agrawal <[email protected]> wrote:
> >> > Is there any condition like all sets should have same no. Of elements
> >> >
> >> > On 10/30/11, SAMMM <[email protected]> wrote:
> >> >> But how does it ensure tht the elements been  removed wouldnot give
> >> >> the same set again ????
> >> >>
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> >> > --
> >> > Sunny Aggrawal
> >> > B.Tech. V year,CSI
> >> > Indian Institute Of Technology,Roorkee
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-- 
Mohit

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