<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> tempVector;
        vector<bool> usedIndex(nums.size());

        backtrack(answer, tempVector, usedIndex, nums);

        sort(answer.begin(), answer.end());
        answer.erase(unique(answer.begin(), answer.end()), answer.end());

        return answer;
    }

    void backtrack(vector<vector<int>>& answer, vector<int>& tempVector,
                   vector<bool>& usedIndex, vector<int>& nums) {
        if (tempVector.size() == nums.size()) {
            answer.push_back(tempVector);
            return;
        }

        for (int i = 0; i < nums.size(); i++) {
            if (usedIndex[i] == true)
                continue;

            tempVector.push_back(nums[i]);
            usedIndex[i] = true;

            backtrack(answer, tempVector, usedIndex, nums);

            tempVector.pop_back();
            usedIndex[i] = false;
        }
    }
};
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