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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