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# Permutations II LeetCode Solution

## Problem – Permutations II LeetCode Solution

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`

## Permutations II LeetCode Solution in Java

``````public class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(nums==null || nums.length==0) return res;
boolean[] used = new boolean[nums.length];
List<Integer> list = new ArrayList<Integer>();
Arrays.sort(nums);
dfs(nums, used, list, res);
return res;
}

public void dfs(int[] nums, boolean[] used, List<Integer> list, List<List<Integer>> res){
if(list.size()==nums.length){
return;
}
for(int i=0;i<nums.length;i++){
if(used[i]) continue;
if(i>0 &&nums[i-1]==nums[i] && !used[i-1]) continue;
used[i]=true;
dfs(nums,used,list,res);
used[i]=false;
list.remove(list.size()-1);
}
}
}
``````

## Permutations II LeetCode Solution in C++

``````class Solution {
public:
void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
if (i == j-1) {
res.push_back(num);
return;
}
for (int k = i; k < j; k++) {
if (i != k && num[i] == num[k]) continue;
swap(num[i], num[k]);
recursion(num, i+1, j, res);
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int> >res;
recursion(num, 0, num.size(), res);
return res;
}
};
``````

## Permutations II LeetCode Solution in Python

``````def permuteUnique(self, nums):
ans = [[]]
for n in nums:
new_ans = []
for l in ans:
for i in xrange(len(l)+1):
new_ans.append(l[:i]+[n]+l[i:])
if i<len(l) and l[i]==n: break              #handles duplication
ans = new_ans
return ans
``````
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