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
>> >>>
>> >>> --
>> >>> You received this message because you are subscribed to the Google
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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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>>
>
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