<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:

    vector<vector<int>> permuteUnique(vector<int>& nums) {

        sort(nums.begin(), nums.end());

        int n = nums.size();

        vector<vector<int>> ans;

        vector<int> t(n);

        vector<bool> vis(n);

        function<void(int)> dfs = [&](int i) {

            if (i == n) {

                ans.emplace_back(t);

                return;

            }

            for (int j = 0; j < n; ++j) {

                if (vis[j] || (j && nums[j] == nums[j - 1] && !vis[j - 1])) {

                    continue;

                }

                t[i] = nums[j];

                vis[j] = true;

                dfs(i + 1);

                vis[j] = false;

            }

        };

        dfs(0);

        return ans;

    }

};
        
   

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