@ross: ya, don't know what i was thinking.!!
On Wed, Jun 22, 2011 at 2:33 PM, ross <jagadish1...@gmail.com> wrote:
> @Harshal,
> Even if you use a buffer of size 256 it is still O(1), because 256 is
> a constant invariant of n...
> Ur solution is correct!
>
>
> On Jun 22, 10:24 am, Harshal <hc4...@gmail.com> wrote:
> > ignore above solution. My bad, did'nt see O(1) space constraint!!
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > On Wed, Jun 22, 2011 at 10:53 AM, Harshal <hc4...@gmail.com> wrote:
> > > You can make use of an auxiliary array(initialized to 0) to store the
> count
> > > of each char and then print it that many times.
> > > char inp[]="abcdaabcdefe";
> > > int buff[256]={0};
> >
> > > for(int i=0;i<strlen(inp);i++)
> > > buff[inp[i]]++;
> >
> > > for(int j=0;j<256;j++)
> > > while(buff[j]--) cout<<(char)j;
> >
> > > On Wed, Jun 22, 2011 at 10:27 AM, Sriganesh Krishnan <
> 2448...@gmail.com>wrote:
> >
> > >> Input will be a string. We need to o/p a string with the order of
> > >> characters same as the input but with same characters grouped
> together.
> > >> I/P: abcdacde
> > >> O/P: aabccdde
> >
> > >> I/P: kapilrajadurga
> > >> O/P: kaaaapilrrjdug
> >
> > >> I/P: 1232
> > >> O/P: 1223 ……………….. O(n) time……….. O(1) space…………….
> >
> > >> how can you approach these type of string related problems, is there
> any
> > >> specific technique involved?
> >
> > >> --
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> >
> > > --
> > > Harshal Choudhary,
> > > Final Year B.Tech CSE,
> > > NIT Surathkal, Karnataka, India.
> >
> > --
> > Harshal Choudhary,
> > Final Year B.Tech CSE,
> > NIT Surathkal, Karnataka, India.
>
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--
Harshal Choudhary,
Final Year B.Tech CSE,
NIT Surathkal, Karnataka, India.
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