@Abhijit. Does your code takes O(N^2)? I think the following code
would do it in O(N)
iterate the string once:
void remove(char *a){
if(!a){
int i = 0, j = 1;
while(a[j]!='\0'){
if(i<0 || a[i]!=a[j]){
if(j-1>i){
a[i+1] = a[j];
}
i++;j++;
}else{
for(;a[j]==a[i];j++);
i--;
}
}
a[i+1] = '\0';
}
}
On Feb 14, 11:17 pm, Tushar Bindal <[email protected]> wrote:
> if we have "RBGGGBGGBR"
> what should be the answer???
> "RBGR" or ""(empty string)
>
> On Mon, Feb 7, 2011 at 3:51 PM, rajan goswami
> <[email protected]>wrote:
>
>
>
>
>
>
>
>
>
> > yeh.
> > Agree with ramkumar.
> > Simplest solution is to use Stack...
>
> > On Sun, Feb 6, 2011 at 8:11 PM, Abhijit K Rao
> > <[email protected]>wrote:
>
> >> I could not get it for recursively, but iteratively, I coded a solution.
> >> If anyone knows recursively,
> >> let us know please.
>
> >> #include<stdio.h>
> >> void main()
> >> {
> >> char s[18]="DGGDBCBHH";
> >> int i=0,j=0;
> >> int count;
> >> while(s[i]!='\0')
> >> {
> >> if(s[i] == s[i+1])
> >> {
> >> count = strlen(s)-2;
> >> while(count--)
> >> {
> >> s[i]=s[i + 2];
> >> i++;
> >> }
> >> s[i]='\0';
> >> i=0;
> >> }
> >> else
> >> {
> >> i++;
> >> }
> >> }
> >> printf("%s",s);
> >> getch();
> >> }
>
> >> I/P: DGGDBCBHH O/P: BCB
>
> >> Best Regards
> >> Abhijit
>
> >> On Sun, Feb 6, 2011 at 1:47 PM, ramkumar santhanam <
> >> [email protected]> wrote:
>
> >>> use stack.
>
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> Tushar Bindal
> Computer Engineering
> Delhi College of Engineering
> Mob: +919818442705
> E-Mail : [email protected], [email protected]
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