<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>

// C++ Solution
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> results;
        unordered_map<int, int> counter;
        for (int num : nums) counter[num]++;
        vector<int> comb;
        backtrack(counter, comb, nums.size(), results);
        return results;
    }
    void backtrack(unordered_map<int, int>& counter, vector<int>& comb, int N,
                   vector<vector<int>>& results) {
        if (comb.size() == N) {
            results.push_back(comb);
            return;
        }
        for (auto& [num, count] : counter) {
            if (count == 0) continue;
            comb.push_back(num);
            count--;
            backtrack(counter, comb, N, results);
            comb.pop_back();
            count++;
        }
    }
};
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