<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:
    void graphWalk(const vector<int>& nums, unordered_map<int, int> counts, vector<vector<int>>& perms, vector<int> perm, const int& num)
    {
        counts[num]--;
        perm.push_back(num);
        if(perm.size() == nums.size())
        {
            perms.push_back(perm);
            return;
        }
        for(const auto & [ num, count ] : counts)
        {
            if(count > 0)
            {
                graphWalk(nums, counts, perms, perm, num);
            }
        }
    }

    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> permutations;
        unordered_map<int, int> num_counts;
        for(auto num : nums)
        {
            if(!num_counts.count(num))
            {
                num_counts[num] = 1;
            }
            else
            {
                num_counts[num]++;
            }
        }
        vector<int> permutation = {};
        for(const auto & [ num, count ] : num_counts)
        {
            graphWalk(nums, num_counts, permutations, permutation, num);
        }
        return permutations;
    }
};
19006