[LC] 47. Permutations II

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

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

Time: O(N!)
Space: O(N)

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        res = []
        if nums is None or len(nums) == 0:
            return res
        self.dfs(nums, 0, res)
        return res
    
    def dfs(self, nums, level, res):
        if level == len(nums):
            res.append(list(nums))
            return
        my_set = set()
        for i in range(level, len(nums)):
            if nums[i] not in my_set:
                my_set.add(nums[i])
                nums[i], nums[level] = nums[level], nums[i]
                self.dfs(nums, level + 1, res)
                nums[i], nums[level] = nums[level], nums[i]

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> arrList = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        boolean[] visited = new boolean[nums.length];
        Arrays.sort(nums);
        helper(arrList, list, visited, nums);
        return arrList;
    }
    
    private void helper(List<List<Integer>> arrList, List<Integer> list, boolean[] visited, int[] nums) {
        if (list.size() == nums.length) {
            arrList.add(new ArrayList<>(list));
            return;
        }   
        for (int i = 0; i < nums.length; i++) {
            if (visited[i] || (i > 0 && nums[i] == nums[i - 1] && !visited[i - 1])) {
                continue;
            }
            visited[i] = true;
            list.add(nums[i]);
            helper(arrList, list, visited, nums);
            list.remove(list.size() - 1);
            visited[i] = false;
        }
    }
}