<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:
    void solve(vector<int> nums, int index, set<vector<int>>& ans) {
        // base case
        if (index >= nums.size()) {
            ans.insert(nums);
            return;
        }

        for (int i = index; i < nums.size(); i++) {
            swap(nums[i], nums[index]);
            solve(nums, index + 1, ans);
        }
    }

public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        set<vector<int>> ans;
        int index = 0;
        solve(nums, index, ans);
        vector<vector<int>> temp(ans.begin(),ans.end());
        // for (auto it = ans.begin(); it != ans.end(); ++it) {
        //     result.push_back(*it);
        // }
        return temp;
    }
};
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