Manu wrote:
> I know counting sort can do in linear time but we dont need to use and
> extra space.
> as we do in case of counting sort.
>
>
> I came up with this sol. but i think it's not stable so if u can find a
> stable algo plzzz....
>
>
> i=0;j=n-1;
>
> while(i<j) {
>      if(a[i]==a[j]) {
>         if(a[i]==0)
>             i++;
>         else
>             j--;     //ie equal to 1
>       continue; //ie go to  while (i<j)
>      }
>     if(a[i]<a[j])
>        i++;       //ie a[i] has to be zero only
>    else {         //ie a[i] =1 and a[j] = 0 therefore swap them
>       swap(a[i],a[j]);
>       i++;
>       j--;
>    }
> } //end of while

Could just do one quicksort-style pivot:

i = 0;
j = n - 1;
while (i < j)
  if (a[i] == 0)
    i++;
  else if (a[j] == 1)
    j--;
  else
    { a[i] = 0; a[j] = 1; }


--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups 
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/algogeeks
-~----------~----~----~----~------~----~------~--~---

Reply via email to