# Permutations II Leetcode Solution

In this post, we are going to solve the Permutations II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

Given a collection of numbers`nums`, that might contain duplicates, return all possible unique permutations in any order.

Example 1:

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

Example 2:

```Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
```

Constraints:

• `1 <= nums.length <= 8`
• `-10 <= nums[i] <= 10`

Now, lets see the leetcode solution of Permutations II Leetcode Solution.

### Permutations II Leetcode Solution in Python

```class Solution:
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
ans = []
used = [False] * len(nums)

def dfs(path: List[int]) -> None:
if len(path) == len(nums):
ans.append(path.copy())
return

for i, num in enumerate(nums):
if used[i]:
continue
if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
continue
used[i] = True
path.append(num)
dfs(path)
path.pop()
used[i] = False

nums.sort()
dfs([])
return ans
```

### Permutations II Leetcode Solutionin CPP

```class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> ans;
sort(begin(nums), end(nums));
dfs(nums, vector<bool>(nums.size()), {}, ans);
return ans;
}

private:
void dfs(const vector<int>& nums, vector<bool>&& used, vector<int>&& path,
vector<vector<int>>& ans) {
if (path.size() == nums.size()) {
ans.push_back(path);
return;
}

for (int i = 0; i < nums.size(); ++i) {
if (used[i])
continue;
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])
continue;
used[i] = true;
path.push_back(nums[i]);
dfs(nums, move(used), move(path), ans);
path.pop_back();
used[i] = false;
}
}
};
```

### Permutations II Leetcode Solution in Java

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

private void dfs(int[] nums, boolean[] used, List<Integer> path, List<List<Integer>> ans) {
if (path.size() == nums.length) {
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] = true;