@Harshal... ur solution is nt correct.....it is nt printing the
characters in order as given in i/p...
i know the solution using auxiliary array and using an extra character
array to hold the o/p string but if any1 knows inplace solution then
plz rply...
On 6/22/11, 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.
>
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