<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 {
private:
    unordered_map<int, int> freq;
    vector<vector<int>> result;
    vector<int> temp;
    int n;
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        n = nums.size();
        temp.resize(n);
        for (auto&& e : nums) freq[e]++;
        set<int> s(nums.begin(), nums.end());
        nums.assign(s.begin(), s.end());
        helper(nums, 0);
        return result;
    }

    void helper(const vector<int>& nums, int pos) {
        if (pos == n) {
            result.emplace_back(temp);
            return;
        }

        for (int i = 0; i < nums.size(); i++) {
            if (freq[nums[i]] > 0) {
                temp[pos] = nums[i];
                freq[nums[i]]--;
                helper(nums, pos + 1);
                freq[nums[i]]++;
            }
        }
    }
};
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