Without the ascii count table as harshal has used above,is it possible to do
the problem in o(n) time?
On Wed, Jun 22, 2011 at 2:57 PM, Harshal <hc4...@gmail.com> wrote:
> @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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--
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD
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