0090. Subsets II

https://leetcode.com/problems/subsets-ii

Description

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1:

**Input:** nums = [1,2,2]
**Output:** [[],[1],[1,2],[1,2,2],[2],[2,2]]

Example 2:

**Input:** nums = [0]
**Output:** [[],[0]]

Constraints:

• 1 <= nums.length <= 10

• -10 <= nums[i] <= 10

ac

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

    private void backtrack(int[] nums, int start, List<List<Integer>> res, List<Integer> note) {
        res.add(new ArrayList<Integer>(note));

        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i-1]) continue;  // skip duplicate
            note.add(nums[i]);
            backtrack(nums, i+1, res, note);
            note.remove(note.size()-1);
        }
    }
}

Sub-optimal with boolean array.

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        boolean[] visited = new boolean[nums.length];

        Arrays.sort(nums);
        process(nums, 0, tmp, visited, res);

        return res;
    }

    private void process(int[] nums, int start, List<Integer> tmp, boolean[] visited, List<List<Integer>> res) {
        res.add(new ArrayList<>(tmp));

        if (start >= nums.length) return;

        for (int i = start; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i-1] && !visited[i-1]) continue; // duplicate
            tmp.add(nums[i]);
            visited[i] = true;
            process(nums, i + 1, tmp, visited, res);
            visited[i] = false;
            tmp.remove(tmp.size() - 1);
        }
    }
}