<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:
    bool vis[8];
    map<vector<int>, bool> M;
    vector<int> V;
    vector<vector<int>> ans;
    int N;

    void dfs(int i, vector<int> num, int digits) {
        vis[i] = 1; digits--;
        V.push_back(num[i]);
        if (digits == 0) {
            if (!M[V]) {
                ans.push_back(V);
                M[V] = 1;
            }

        } else {
            for (int i = 0; i < N; i++) {
                if (!vis[i]) {
                    dfs(i, num, digits);
                }
            }
        }

        V.pop_back();
        vis[i] = 0;
    }

    vector<vector<int>> permuteUnique(vector<int>& nums) {
        N = nums.size();
        for (int i = 0; i < N; i++) dfs(i, nums, N);
        return ans;
    }
};
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