<p>Given an integer array <code>nums</code> of <strong>unique</strong> elements, return <em>all possible</em> <span data-keyword="subset"><em>subsets</em></span> <em>(the power set)</em>.</p>

<p>The solution set <strong>must not</strong> contain duplicate subsets. Return the solution in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0]
<strong>Output:</strong> [[],[0]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>All the numbers of&nbsp;<code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> result;
    int n;
    unordered_set<int> st;
    void solve(int i, vector<int>& temp, vector<int>& nums) {
        if (i >= n) {
            result.push_back(temp);
            return;
        }
        // for(int i = index ; i<n; i++){
        if (st.find(nums[i]) == st.end()) {
            temp.push_back(nums[i]);
            st.insert(nums[i]);
            solve(i + 1, temp, nums);
            temp.pop_back();
            st.erase(nums[i]);
            solve(i + 1, temp, nums);
        }
        // }
    }

    vector<vector<int>> subsets(vector<int>& nums) {

        vector<int> temp;
        n = nums.size();
        solve(0, temp, nums);
        return result;
    }
};
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