Odtnhj

Given a set of distinct integers, I want to find all the possible subsets (for [1,2,3], the code should print [1], [1,2], [1,3], [1,2,3], [2], [2,3], [3] (not necessarily in that order).

There are a few solutions (like this one) out there but what I want to do is to re-implement the bellow solution with a new recursion and no for loop by passing around the indexes: (start = 0)

public void forSolution(List<List<Integer>> res, int nums, List<Integer> list, int start) 

 for (int i = start; i < nums.length; i++) 
 List<Integer> tmp = new ArrayList<>(list);
 tmp.add(nums[i]);
 res.add(new ArrayList<>(tmp));
 forSolution(res, nums, tmp, i + 1);
 

I thought I need to pass two integers to the method, one for keeping the record of index and the other one for keeping the start point, but I am having problem on when I need to do the index increment (vs start increment).
Any help would be appreciated.

I think the algorithm gets easier if you don't bother with indices.

The basic idea is that for any given sublist, each element of the original list is either included or not included. The list of all possible sublists is simply all possible combinations of including / not including each element.

For a recursive implementation, we can consider two cases:

1. The input list is empty. The empty list only has a single sublist, which is the empty list itself.


2. 
   

   The input list consists of a first element x and a list of remaining elements rest. Here we can call our function recursively to get a list of all sublists of rest. To implement the idea of both including and not including x in our results, we return a list consisting of

   
   
   
   • each element of sublists(rest) with x added at the front, representing all sublists of our original list that contain x, and
   
   
   • each element of sublists(rest) as is (without x), representing all sublists of our original list that don't contain x.
   
   

For example, if the list is [1, 2, 3], we have x = 1 and rest = [2, 3]. The recursive call sublists(rest) produces [2, 3], [2], [3], . For each of those sublists we

• prepend x (which is 1), giving [1, 2, 3], [1, 2], [1, 3], [1], and


• don't prepend x, giving [2, 3], [2], [3], .

Concatenating those parts gives our total result as [1, 2, 3], [1, 2], [1, 3], [1], [2, 3], [2], [3], .

Sample implementation:

use strict;
use warnings;

sub sublists 
 if (!@_) 
 return ;
 
 my $x = shift @_;
 my @r = sublists(@_);
 return (map [$x, @$_], @r), @r;


for my $sublist (sublists 1, 2, 3) 
 print "[" . join(", ", @$sublist) . "]n";

Output:

[1, 2, 3]
[1, 2]
[1, 3]
[1]
[2, 3]
[2]
[3]