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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