@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
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