<p>Given a collection of numbers, <code>nums</code>,&nbsp;that might contain duplicates, return <em>all possible unique permutations <strong>in any order</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,2]
<strong>Output:</strong>
[[1,1,2],
 [1,2,1],
 [2,1,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 8</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
</ul>

class Solution {
public:
void solve(int idx, int n, vector<int> &nums, vector<vector<int>> &ans) {
        if(idx >= n) {
            ans.push_back(nums);
            return;
        }

        for(int i=idx; i<n; i++) {
            swap(nums[i], nums[idx]);
            solve(idx+1, n, nums, ans);
            swap(nums[i], nums[idx]);
        }
    }

    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> ans;
        int n = nums.size();
        solve(0, n, nums, ans);
        set<vector<int>>s;
        for(auto x:ans) s.insert(x);
        ans.clear();
        for(auto x:s) ans.push_back(x);
        return ans;
    }
};


    

    
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