as an eg.
let ab be the string, and 3 characters length string is wht is expected..
a - -
b - -
a a -
a b -
b a -
b b -
a a a
a a b
a b a
a b b
b a a
b a b
b b a
b b b
On Fri, Aug 5, 2011 at 12:49 PM, Gaurav Menghani
<gaurav.mengh...@gmail.com>wrote:
> The basic idea is that for every position of the string, you fill it
> with all possible alphabets in your set of allowed alphabets, let the
> set be called alphabet.
>
> Now, you can do this recursively.
> backtrack(s,l) denotes, a string s has been formed, and is of length
> l. Now we need to add more letters to it. If l is equal to the maximum
> length, then you just print the string s, and return.
>
> Otherwise, append the characters available to you. For example, if in
> the current scenario, the call is backtrack("po",2), and alphabet =
> {'o','p'} and maxlen=3, then we can append 'o' and 'p' to the string
> "po", and hence call backtrack("poo",3) and backtrack("pop",3).
>
> You start the process by calling backtrack("",0), and setting maxlen
> and alphabet to the appropriate values.
>
> 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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