stack for sure..
pop out the enteries till they r same and push this entry back once.

On Wed, Feb 16, 2011 at 11:18 AM, jalaj jaiswal
<jalaj.jaiswa...@gmail.com>wrote:

> can any1 tell me the question .. m gettin half of the post ,...
>
> On Wed, Feb 16, 2011 at 10:41 AM, Jammy <xujiayiy...@gmail.com> 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 <tushicom...@gmail.com> 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 <
>> rajan.goswam...@gmail.com>wrote:
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > > yeh.
>> > > Agree with ramkumar.
>> > > Simplest solution is to use Stack...
>> >
>> > > On Sun, Feb 6, 2011 at 8:11 PM, Abhijit K Rao <
>> abhijitkrao...@gmail.com>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 <
>> > >> ramkumars....@gmail.com> wrote:
>> >
>> > >>> use stack.
>> >
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>> > --
>> > Tushar Bindal
>> > Computer Engineering
>> > Delhi College of Engineering
>> > Mob: +919818442705
>> > E-Mail : tusharbin...@jugadengg.com, tushicom...@gmail.com
>>
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>>
>
>
> --
> With Regards,
> *Jalaj Jaiswal* (+919019947895)
> Software developer, Cisco Systems
> B.Tech IIIT ALLAHABAD
>
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