Below is a solution -
#include <stdio.h>
int main(int argc, char* argv[]){
int i = 0;
char *array = argv[1];
char prev = array[0];
while(array[i]){
int count = 0;
while(prev == array[i]){
i +=1;
count++;
}
printf("%d%c,",count,prev);
prev = array[i];
}
printf("\n");
}
Regards
Abhijeet Srivastva
On Wed, Aug 10, 2011 at 9:04 AM, monish001 <monish.gup...@gmail.com> wrote:
> Given : ddaaaabbccae
> O/P : 2d4a2b2c1a1e
> What is happening? What algo?
>
> Thanks and regards
> Monish
>
> On Aug 9, 5:59 pm, ankit sambyal <ankitsamb...@gmail.com> wrote:
> > Given an array of characters, change the array to something as shown in
> the
> > following example.
> > Given : ddaaaabbccae
> > O/P : 2d4a2b2c1a1e
> >
> > Do it in the most efficient manner both in terms of time and space ...
>
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