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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