Or one could just simulate a counting from 0 to (numchars^N)-1 in base
numchars.
.......
code:
void printit(int N,char chars[],int index[]){
for(int i=0;i<N;i++){
printf("%c",chars[index[i]]);
}
printf("\n");
}
void generate(int numchars,char chars[],int N){
int index[100]={0};
int allmax=0;
int maxdigit=numchars-1;
printit(N,chars,index);
while(!allmax){
// add one;
index[0]++;
allmax=0;
for(int i=0;i<N;i++){
if(index[i]>=numchars){
index[i]=0; index[i+1]++;
}
if(i==0&&index[i]==maxdigit){
allmax=1;
}
allmax = (index[i]==maxdigit)?allmax&1:0;
}
printit(N,chars,index);
}
}
int main(){
char chars [] ={'p','o'};
int numchars =sizeof(chars)/sizeof(char);
int N=3;
generate(numchars,chars,N);
}
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
> >>
> >> --
> >> You received this message because you are subscribed to the Google
> Groups
> >> "Algorithm Geeks" group.
> >> To post to this group, send email to algogeeks@googlegroups.com.
> >> To unsubscribe from this group, send email to
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> >> For more options, visit this group at
> >> http://groups.google.com/group/algogeeks?hl=en.
> >>
> >
> >
> >
> > --
> > Regards,
> > Kamakshi
> > kamakshi...@gmail.com
> >
> > --
> > You received this message because you are subscribed to the Google Groups
> > "Algorithm Geeks" group.
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> >
>
>
>
> --
> Gaurav Menghani
>
> --
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