Find Subsets or Power set

Solution to return all possible subsets (the power set) problem.

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        int n = nums.length;
        int twoPowN = 1 << n;
        List<List<Integer>> r = new ArrayList<>();
        
        for (int i=0;i<twoPowN;i++) {
            List<Integer> sr = new ArrayList<>();
            for(int j=0;j<n;j++) {
                int bits = 1 << j;
                if ((i & bits) > 0)
                    sr.add(nums[j]);
            }
            r.add(sr);
        }
        
        return r;
    }
    
}

The approach used is explained in this geeksforgeeks link.

Alternate approach using Recursion

This approach is a direct port of the python code provided in the leetcode discussion.

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        List<Integer> path = new ArrayList<Integer>();

        dfs(nums, 0, path, ans);
        return ans;
    }
    
    private void dfs(int[] nums, int ndx, List<Integer> path, List<List<Integer>> ans)     {
        ans.add(path);

        for(int i=ndx;i<nums.length;i++){
            List<Integer> npath = new ArrayList<Integer>(path);
            npath.add(nums[i]);
            ndx = ndx + 1;
            dfs(nums, ndx, npath, ans);
        }
    }
    
}