<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:
    void recursion(int sz, vector<vector<int>>&ans, vector<int>&cp, vector<int>& r)
    {
        if(r.size() == sz)
        {
            ans.push_back(r);
            return;
        }
        for(int i = 0;i<cp.size();i++)
        {
            r.push_back(cp[i]);
            vector<int> copy = cp;
            copy.erase(copy.begin(), copy.begin()+i+1);
            recursion(sz, ans, copy, r);
            r.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> ans;
        ans.push_back({});
        vector<int> r;
        for(int i = 1;i<=nums.size();i++)
        {
            recursion(i, ans, nums, r);
        }
        return ans;
    }
};
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