Permutations II

Problem

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example, [1,1,2] have the following unique permutations:

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

Solution

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

    private void helper(List<List<Integer>> res, List<Integer> list, int[] nums, boolean[] used) {
        if (list.size() == nums.length) {
            res.add(new ArrayList<>(list));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (used[i]) continue;
            if (i > 0 && nums[i] == nums[i-1] && used[i-1]) continue; //!used[i-1] works too
            list.add(nums[i]);
            used[i] = true;
            helper(res, list, nums, used);
            list.remove(list.size() - 1);
            used[i] = false;
        }
    }
}

Analysis

This is a variation of Permutation Problem
The difference in this problem is that given input nums can have duplicated elements
Therefore, we first sort the nums (typical solution when duplicated elements are in array)
Then we use a boolean[] used to store the status of using each number in nums
We don't have a start parameter in helper() function but we pass the boolean[] used
Then in the for statement we loop from i = 0 to the end
We continue when we face the following conditions:

• When we already used that number in helper() function (used[i] == true)
• When we have same consecutive elements and the former one has already used before (nums[i] == num[i-1] && used[i-1])