Merging Arrays and Grouping

Question

Hi guys,
I am trying to write an algorithm to merge an array of objects to cut down its size. Easier to understand by example.
Lets say I have 3 numbers , 1, 2 and 3. And 3 elements. So I want to create all combination's of these numbers. Like :
1,1,1
1,1,2
1,1,3
1,2,1
1,2,2
1,2,3
1,3,1
etc..
Then lets say I only had removed some of the combinations and was left with :
1,1,1
1,1,2
1,1,3
1,2,1
1,2,2
1,3,1
1,3,2
1,3,3
I would want this shortened to :
1,1,=
1,2,1
1,2,2
1,3,=
(= meaning all possible numbers, in this case 1 2 and 3)
Also if I had :
1,1,1
1,1,2
1,1,3
1,2,1
1,2,2
1,2,3
1,3,1
1,3,2
1,3,3
2,1,1
This would be shortened to :
1,=,=
2,1,1
And if I had :
1,1,1
1,1,2
1,2,1
1,2,3
1,3,1
This would be shortened to :
1,=,1
1,1,2
1,2,3
I think you'll get the idea from that. There could be more that 3 elements. I have been trying to work this out but its doing my head in.
Would be grateful if anyone could give me and tips or point me in the right direction.
Thanks.
Edited by: joeJack on Mar 1, 2010 12:21 PM
Changed from asterisk to equals sign as asterisk was creating bold text.

Answer 1

Hi
My suggested Algorithm is: Take a LinkedHashSet and put all possible combinations of merged objects from the number of elements. From your example all possible combination is 1==,2==,3==,11=,12=,13=,21=,22=,23=,31=,32=,33=. with all values 0(zero). Keep in mind that most general combination will be kept first like,1== and most specific at last i.e 12=,13=...etc
Now scan the each element one by on from array and if first character is 1 of the element increase the count of 1== by 1, if second character is 2 i.e element of the form 12 then increase the count of 12= by 1..and so on till last -1 character of the element.So that way we calculate the frequency of each element in on go.
Now check the number of the count for each element. Say for 1== count is 5 and we know that for 1== count has to be 9 (from permutation possible value 3* 3=9). Similarly go and check each element count and those element count which match the possible permutation value criteria they will be replace by that combination.
if array sixe is >>> than no of character in one string than this algorith will run ~~ n time where n size of array.