I am just increasing the size of the current string by one. So that a
new character can be appended.
On Sat, Aug 6, 2011 at 11:01 AM, Tushar Bindal <tushicom...@gmail.com> wrote:
> @gaurav
> didn't get this:
> Just to increase the size of the string by one.
>
> Then you can put any character at the the new last position, which is 'l'.
>
> can u pls explain that?
>
> On Fri, Aug 5, 2011 at 2:57 PM, Nitin Nizhawan <nitin.nizha...@gmail.com>
> wrote:
>>
>> Ok, Thanks
>>
>> On Fri, Aug 5, 2011 at 2:53 PM, Gaurav Menghani
>> <gaurav.mengh...@gmail.com> wrote:
>>>
>>> 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
>>> >> >>> Groups
>>> >> >>> "Algorithm Geeks" group.
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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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>>>
>>>
>>>
>>> --
>>> Gaurav Menghani
>>>
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>>
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>
>
>
> --
> Tushar Bindal
> Computer Engineering
> Delhi College of Engineering
> Mob: +919818442705
> E-Mail : tushicom...@gmail.com
> Website: www.jugadengg.com
>
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