A solution in java without arrays...
class Combinations
{
static ArrayList<String> words = new ArrayList<String>();
static void getCombinations(ArrayList<String> words)
{
for(int j=0;j<words.length;word++)
{
//find the index of the word
String word = words.get(j);
//for each words, get the combinations for words AFTER IT
for (int i=0;i<word.length;word++){
//iterate thru the reamining words
for(int m=j;m<words.size()-1;m++){
//get the next word
String combWord =words.get(m);
//now combine it with current 'LETTER' of this WORD
for(int k=0;k<combWord.length;k++)
{
System.out.println(word.charAt(i)+combWord.charAt(k));
}
}
}
}
}
public static void main (String args[])
{
words.add("ace");
words.add("bowl");
words.add("put");
getCombinations(words);
}
}
On Monday, October 1, 2012 8:47:48 AM UTC-7, ((** VICKY **)) wrote:
> No we don't need to care about repeated strings. :) Thanks for the
> response folks.
>
> On Mon, Oct 1, 2012 at 8:42 PM, saurabh agrawal
> <[email protected]<javascript:>
> > wrote:
>
>> Do we need to handle cases when the same string will appear again??
>> In that case we can sort individual array and remove duplicates.
>>
>>
>> On Mon, Oct 1, 2012 at 9:54 AM, Rahul Singh <[email protected]<javascript:>
>> > wrote:
>>
>>> check this out..
>>>
>>> #include<iostream>
>>> #include<stdlib.h>
>>> using namespace std;
>>>
>>> void print_sets(string *s,int pos,int n,char *to_print)
>>> {
>>> if(pos==n)
>>> {
>>> return;
>>> }
>>>
>>> for(int i=0;i<s[pos].length();i++)
>>> {
>>> to_print[pos] = s[pos][i];
>>> print_sets(s,pos+1,n,to_print);
>>> if(pos==n-1)
>>> {
>>> for(int j=0;j<n;j++)cout<<to_print[j];
>>> cout<<endl;
>>> }
>>> }
>>> return;
>>> }
>>> int main()
>>> {
>>> int n;
>>> cin>>n;
>>> string s[n];
>>>
>>> for(int i=0;i<n;i++)
>>> {
>>> cin>>s[i];
>>> }
>>> char *to_print = new char[n];
>>> print_sets(s,0,n,to_print);
>>> }
>>>
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>>
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>
>
>
> --
> Cheers,
>
> Vicky
>
>
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