@Dave,
Can you please explain the algo? It's getting very difficult to understand
the code ..
On 3 August 2011 01:14, Dave <dave_and_da...@juno.com> wrote:
> @Pankaj: Assuming generously that by N^2 you mean N*N instead of N
> exclusive-or 2, your very first statement is already O(N^2), as it
> will take that long just to set the array to zero.
>
> Here is a radix sort to sort an array x[N] containing values from 1 to
> N*N in O(N):
>
> int a[N], b[N], i;
> // initialize and tally occurrences of first radix-N digit of x[i]-1:
> for( i = 0 ; i < N ; ++i )
> a[i] = 0;
> for( i = 0 ; i < N ; ++i )
> a[(x[i]-1)/N]++;
> // compute starting point for each radix digit:
> a[N-1] = N - a[N-1];
> for( i = N-2 ; N >= 0 ; --i )
> a[i] = a[i+1] - a[i];
> // move numbers from array x to temp array b:
> for( i = 0 ; i < N ; ++i )
> b[a[(x[i]-1)/N]++] = x[i];
>
> // initialize and tally occurrences of second radix-N digit of x[i]-1:
> for( i = 0 ; i < N ; ++i )
> a[i] = 0;
> for( i = 0 ; i < N ; ++i )
> a[(x[i]-1)%N]++;
> // compute starting point for each radix digit:
> a[N-1] = N - a[N-1];
> for( i = N-2 ; N >= 0 ; --i )
> a[i] = a[i+1] - a[i];
> // move numbers from temp array b back to array x:
> for( i = 0 ; i < N ; ++i )
> x[a[(x[i]-1)%N]++] = b[i];
> // array is now sorted. Run time is O(N). Space is O(N).
>
> Dave
>
> On Aug 2, 11:04 am, pankaj kumar <pancsen...@gmail.com> wrote:
> > int a[N^2]={0},i,j;
> > for(i=0;i<N^2;i++)
> > {
> > cin>>j;
> > a[j]++;
> >
> > }
> >
> > for(i=0;i<N^2;i++)
> > {
> > if(a[i]!=0)
> > {
> > while(a[i]--)
> > {
> > cout<<i<<"\t";
> >
> >
> >
> > }
> > }- Hide quoted text -
> >
> > - Show quoted text -
>
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