<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 {
    int permLength;
    set<vector<int>> permutations;

    void permuteRecurse(vector<int>& curNums, map<int, int>& numFreq) {
        if (curNums.size() == permLength) {
            permutations.insert(curNums);
            return;
        }

        vector<int> firstNums;
        for (auto const& it : numFreq) {
            for (int i = 0; i < it.second; i++) {
                firstNums.push_back(it.first);
            }
        }

        for (int n : firstNums) {
            numFreq[n]--;
            curNums.push_back(n);

            permuteRecurse(curNums, numFreq);

            numFreq[n]++;
            curNums.pop_back();
        }
    }

   public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        permutations = {};
        permLength = nums.size();

        vector<int> curNums;
        map<int, int> numFreq;

        for (int n : nums) {
            numFreq[n]++;
        }

        permuteRecurse(curNums, numFreq);
        return vector<vector<int>>(permutations.begin(), permutations.end());
    }
};
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