<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) {
        vector<vector<int>> answer;
        vector<int> currentComb;
        sort(nums.begin(), nums.end());
        backtracking(nums, answer, currentComb, 0);
        return answer;
    }

    void backtracking(vector<int>& nums, vector<vector<int>>& answer, vector<int> currentComb, int current) {
        if (currentComb.size() == nums.size()) {
            answer.push_back(currentComb);
            return;
        }

        for(int i=0; i<nums.size(); i++) {
            if (i>0 && nums[i] == nums[i-1]) {
                continue;
            }
            currentComb.push_back(nums[i]);
            if (checkFrequency(nums, currentComb)) {
                backtracking(nums, answer, currentComb, i);
            }
            currentComb.pop_back();
        }
    }

    bool checkFrequency(vector<int>& nums, vector<int>& current) {
        unordered_map<int, int> count;
        for(auto itr: nums) {
            count[itr] += 1;
        }

        for(auto itr: current) {
            if(!count[itr]) {
                return false;
            }
            count[itr]--;
        }

        return true;
    }
};
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