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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