Ok, Thanks

On Fri, Aug 5, 2011 at 2:53 PM, Gaurav Menghani
<gaurav.mengh...@gmail.com>wrote:

> Even if the number of elements is more than two, it is possible with
> bitwise operations, but it gets clumsy.
>
> Suppose your alphabet has 4 characters. You can either:
> - Count from 0 to (1<<4*n)-1 and use four bits to denote the selection
> of the alphabet. Also, only one bit amongst those four should be set.
> It is highly inefficient.
> - Keep n nested loops and inside each loop you iterate from 0 to
> (1<<4)-1 and use the standard bitwise operations. The con here is that
> you have to hardcode the number of nested loops.
>
> On Fri, Aug 5, 2011 at 2:44 PM, Nitin Nizhawan <nitin.nizha...@gmail.com>
> wrote:
> > @Varun  I think it can be done using bits, if input character set has
> only
> > two elements. Or could u plz explain?
> >
> > On Fri, Aug 5, 2011 at 2:29 PM, Varun Jakhoria <varunjakho...@gmail.com>
> > wrote:
> >>
> >> I think it can be done using bitwise ANDing with a mask
> >>
> >> On Fri, Aug 5, 2011 at 12:58 PM, Gaurav Menghani
> >> <gaurav.mengh...@gmail.com> wrote:
> >> > An Implementation:
> >> >
> >> > #include<iostream>
> >> > #include<string>
> >> > using namespace std;
> >> >
> >> > string alphabet;
> >> > int maxlen;
> >> > void backtrack(string s,int l)
> >> > {
> >> >  if(l==maxlen) { cout<<s<<endl; return; }
> >> >  s.push_back('-');
> >> >  for(int i=0;i<alphabet.size();i++)
> >> >        { s[l]=alphabet[i]; backtrack(s,l+1); }
> >> > }
> >> >
> >> > int main()
> >> > {
> >> >  maxlen=3;
> >> >  alphabet="op";
> >> >  backtrack("",0);
> >> >  return 0;
> >> > }
> >> >
> >> >
> >> > On Fri, Aug 5, 2011 at 12:42 PM, Kamakshii Aggarwal
> >> > <kamakshi...@gmail.com> wrote:
> >> >> @gaurav:i could not understand ur sol.can u explain it again..
> >> >>
> >> >> On Fri, Aug 5, 2011 at 12:32 PM, Gaurav Menghani
> >> >> <gaurav.mengh...@gmail.com>
> >> >> wrote:
> >> >>>
> >> >>> On Fri, Aug 5, 2011 at 12:20 PM, Kamakshii Aggarwal
> >> >>> <kamakshi...@gmail.com> wrote:
> >> >>> > given a set of letters and a length N, produce all possible
> >> >>> > output.(Not
> >> >>> > permutation). For example, give the letter (p,o) and length of 3,
> >> >>> > produce
> >> >>> > the following output(in any order you want, not just my example
> >> >>> > order)
> >> >>> >
> >> >>> > ppp ppo poo pop opp opo oop ooo
> >> >>> >
> >> >>> > another example would be given (a,b) and length 2
> >> >>> >
> >> >>> > answer: ab aa bb ba
> >> >>> >
> >> >>> > --
> >> >>> > Regards,
> >> >>> > Kamakshi
> >> >>> > kamakshi...@gmail.com
> >> >>>
> >> >>> This can be done easily by backtracking
> >> >>>
> >> >>> void backtrack(string s, int l)
> >> >>> {
> >> >>>   if(l == maxlen) { cout<<s<<endl; return; }
> >> >>>
> >> >>>   s.push_back('-');
> >> >>>   for(int i=0;i<alphabet.size();i++)
> >> >>>   {
> >> >>>     s[l]=alphabet[i];
> >> >>>     backtrack(s,l+1);
> >> >>>   }
> >> >>> }
> >> >>>
> >> >>> --
> >> >>> Gaurav Menghani
> >> >>>
> >> >>> --
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> >> >>>
> >> >>
> >> >>
> >> >>
> >> >> --
> >> >> Regards,
> >> >> Kamakshi
> >> >> kamakshi...@gmail.com
> >> >>
> >> >> --
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> >> >>
> >> >
> >> >
> >> >
> >> > --
> >> > Gaurav Menghani
> >> >
> >> > --
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> >> >
> >>
> >>
> >>
> >> --
> >> Varun Jakhoria
> >> ...it's only about 0's & 1's
> >>
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>
>
> --
> Gaurav Menghani
>
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