<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 {
    vector<int> nextGreaterPermutation(vector<int> &A) {
        int n = A.size();

        int ind = -1; 
        for (int i = n - 2; i >= 0; i--) {
            if (A[i] < A[i + 1]) {
                ind = i;
                break;
            }
        }

        if (ind == -1) {
            reverse(A.begin(), A.end());
            return A;
        }

        for (int i = n - 1; i > ind; i--) {
            if (A[i] > A[ind]) {
                swap(A[i], A[ind]);
                break;
            }
        }

        reverse(A.begin() + ind + 1, A.end());

        return A;
    }
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> ans;
        int n = nums.size();
        set<vector<int>> st;
        int x = n;
        int count = 1;
        while(x){
            count = count*x;
            x--;
        }

        for(int i=0; i<count; i++){
            vector<int> temp = nextGreaterPermutation(nums);
            if(st.find(temp) != st.end()){
                continue;
            }
            ans.push_back(temp);
            st.insert(temp);
        }

        return ans;
    }
};
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