@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!!
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> 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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