<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>> output;

    int n, k;

    void backtrack(int first, vector<int> curr, vector<int>& nums) {

        // if the combination is done

        if (curr.size() == k) output.push_back(curr);

        for (int i = first; i < n; ++i) {

            // add i into the current combination

            curr.push_back(nums[i]);

            // use next integers to complete the combination

            backtrack(i + 1, curr, nums);

            // backtrack

            curr.pop_back();

        }

    }

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

        n = nums.size();

        for (k = 0; k < n + 1; ++k) {

            vector<int> curr;

            backtrack(0, curr, nums);

        }

        return output;

    }

};
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