<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<int> v;
        vector<vector<int>> ans;
        unordered_set<int> indices;
        sort(nums.begin(), nums.end());

        for(int i=0; i<nums.size(); i++) {
            if(i>0 && nums[i] == nums[i-1])
                continue;
            indices.insert(i);
            constructPermutations(nums, i, ans, v, indices);
            indices.erase(i);
        }

        return ans;
    }

    void constructPermutations(vector<int> &nums, int index, vector<vector<int>> &ans, vector<int> v, unordered_set<int> indices) {
        v.push_back(nums[index]); //0
        if(v.size() == nums.size()) {
            ans.push_back(v);
            return;
        }
        for(int i=0; i<nums.size(); i++) {
            if(indices.contains(i))
                continue;

            if(i>0 && !indices.contains(i) && !indices.contains(i-1) && nums[i]==nums[i-1])
                continue;
            
            indices.insert(i);
            constructPermutations(nums, i, ans, v, indices);
            indices.erase(i);
        }
    }
};
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