<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>

void recurse(map<vector<int>,int> & permutations , vector<int>& numsarr,vector<int>& nums){
    if(nums.empty()) permutations[numsarr] = 1;
    vector<int> nums3 = nums;
    for(int j = 0 ; j < nums.size();j++){
            int num = nums[j];
            numsarr.push_back(num);
            nums.erase(nums.begin() + j);
            recurse(permutations,numsarr,nums);
            numsarr.pop_back();
            nums = nums3;


    }
   
}


class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        map<vector<int>,int>  permutations ;
        vector<int> numsarr;
        vector<vector<int>> perm;
        recurse(permutations,numsarr,nums);
        for(auto iter:permutations) perm.push_back(iter.first);
        return perm;
    }
};
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