Subsets II

Problem

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,

If nums = [1,2,2], a solution is:

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

Solution

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        helper(res, new ArrayList<>(), nums, 0);
        return res;
    }

    private void helper(List<List<Integer>> res, List<Integer> list, int[] nums, int start) {
        res.add(new ArrayList<>(list));
        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i-1]) continue; // i > start used to avoid nums[0] == nums[-1] error 
            list.add(nums[i]);
            helper(res, list, nums, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

Analysis

This is a general way to get all subsets of a given array. The difference of this solution to 78 Subsets is that the given array might contain duplicated elements. I first thought of using set to do this problem. However, it's hard to check equality of a list (which is an object). For example, a list {2, 1, 2} is different from {1,2,2} because of orders. Hence, the correct and easiest way to avoid duplicates in this backtracking solution is that, we first sort the given array and check if nums[i] == nums[i-1]. If true, we just continue the loop. Besides these, everything is same as getting subsets from an array with distinct elements.