<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) {
        const int n = nums.size();
        unordered_set<string> seen;
        vector<int> curr;
        vector<vector<int>> res;
        unordered_set<int> used;
        function<void(int)> helper = [&](int i) {
            //if (used.count(i)) return;
            used.insert(i);
            curr.push_back(nums[i]);
            if (used.size() == n) {
                string s;
                for (int v : curr) {
                    s.append(":"+to_string(v));
                }
                if (!seen.count(s)) {
                    seen.insert(s);
                    res.push_back(curr);
                }
            }
            for (int i = 0; i < n; ++i) {
                if (!used.count(i))
                    helper(i);
            }
            curr.pop_back();
            used.erase(i);
        };
        for (int i = 0; i < n; ++i) {
            helper(i);
        }
        return res;
    }
};

/*
1   1   1

0   1   2
0   2   1

sort

1   1   2

0   1   2
0   2   1
1   2   0


*/
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