<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>

auto init = []() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  return 0;
}();

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
  void calcSubset(vector<int> &nums, set<vector<int>> &A, vector<int> &curSet, int index) {
    A.insert(curSet);
    for (size_t i = index; i < nums.size(); i++) {
      curSet.push_back(nums[i]);
      calcSubset(nums, A, curSet, i + 1);
      curSet.pop_back();
    }
  }
  vector<vector<int>> subsets(vector<int> &nums) {
    set<vector<int>> A;
    vector<int> curSet;
    calcSubset(nums, A, curSet, 0);
    vector<vector<int>> p(A.begin(), A.end());
    return p;
  }
};
9000