Solution 1:
need two scan :
start from left and replace one by one by non-space character ( thus
have minimal replacemnt)
count =0;
for (int i=0;i<len(str);i++)
{
if(str[i] !=NONSPCECHAR)
{ str[count++]=str[i]
}
}
ps: any way it needs Left shift...and all solution must need left
shift if you try to compact it in lower index area.
On 10/11/11, abhishek sharma <abhishek.p...@gmail.com> wrote:
> Can in place compaction be done without left shifts?
>
>
>
> --
> Nice Day
>
> Abhishek Sharma
> Bachelor of Technology
> IIT Kanpur (2009)
>
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**DIPANKAR DUTTA
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