@mohit: that will modify the original array
and also time =O(nlogn)...
On Mon, Sep 5, 2011 at 1:01 AM, Ankuj Gupta <ankuj2...@gmail.com> wrote:
> @mohit: that will modify the original array
>
> On Sep 4, 6:40 pm, sarath prasath <prasathsar...@gmail.com> wrote:
> > here is my approach
> > where i left the non repeating characters as it is and done some good
> code..
> > char * runlengthencode(char* str,int size)
> > {
> > int i,j,flag=0;
> > for(i=0,j=1;str[i]&&str[j]&&j<size;i++,j++)
> > {
> > while(str[i]==str[j])
> > {
> > j++;
> > flag=1;
> >
> > }
> > if(flag)
> > {
> > j=j-1;
> > str[i+1]=48+(j-i+1);
> > flag=0;
> > i=j;
> > j++;
> > }
> > }
> > return str;
> >
> >
> >
> >
> >
> >
> >
> > }
> > On Sat, Sep 3, 2011 at 6:54 PM, Aman Kumar <amanas...@gmail.com> wrote:
> > > Hiii
> > > if array is given like this
> >
> > > arr[]=aabcabbcdeadef
> >
> > > convert this array into like
> >
> > > arr[]=a4b3c2d2e2f1
> >
> > > how can we do this
> >
> > > can we do it with space complexity O(1).
> >
> > > reply asap
> >
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