Great.
On Fri, Aug 5, 2011 at 2:42 PM, Kamakshii Aggarwal
<kamakshi...@gmail.com> wrote:
> @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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>>>> >> "Algorithm Geeks" group.
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>>>> >>
>>>> >
>>>> >
>>>> >
>>>> > --
>>>> > Regards,
>>>> > Kamakshi
>>>> > kamakshi...@gmail.com
>>>> >
>>>> > --
>>>> > You received this message because you are subscribed to the Google
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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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>
--
Gaurav Menghani
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