@gaurav:i got it..thanks for the solution

On Fri, Aug 5, 2011 at 2:34 PM, Kamakshii Aggarwal <kamakshi...@gmail.com>wrote:

> @gaurav:can u please explain what is the purpose of this line..
> s.push_back('-');
>
>
> On Fri, Aug 5, 2011 at 1:10 PM, Nitin Nizhawan 
> <nitin.nizha...@gmail.com>wrote:
>
>> 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
>>> >>
>>> >> --
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>>> >>
>>> >
>>> >
>>> >
>>> > --
>>> > Regards,
>>> > Kamakshi
>>> > kamakshi...@gmail.com
>>> >
>>> > --
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>>> >
>>>
>>>
>>>
>>> --
>>> Gaurav Menghani
>>>
>>> --
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>>>
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>>
>
>
>
> --
> Regards,
> Kamakshi
> kamakshi...@gmail.com
>



-- 
Regards,
Kamakshi
kamakshi...@gmail.com

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