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 > <saura...@gmail.com<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 <riit...@gmail.com<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); >>> } >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To post to this group, send email to algo...@googlegroups.com<javascript:> >>> . >>> To unsubscribe from this group, send email to >>> algogeeks+...@googlegroups.com <javascript:>. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algo...@googlegroups.com<javascript:> >> . >> To unsubscribe from this group, send email to >> algogeeks+...@googlegroups.com <javascript:>. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > > > -- > Cheers, > > Vicky > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/MbGDQZ6k2NEJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.