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 [email protected]
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/algogeeks
-~----------~----~----~----~------~----~------~--~---
