can any1 tell me the question .. m gettin half of the post ,...
On Wed, Feb 16, 2011 at 10:41 AM, Jammy <[email protected]> wrote:
> @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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>
--
With Regards,
*Jalaj Jaiswal* (+919019947895)
Software developer, Cisco Systems
B.Tech IIIT ALLAHABAD
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