<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 recans(int st, vector<int>& a, set<vector<int>>& s, vector<int>& nums) {
        // Base case: if index exceeds size, stop the recursion
        if (st == nums.size()) {
            s.insert(a);
            return;
        }

        // Insert current subset into set
        s.insert(a);

        // Recursive call including the current element
        a.push_back(nums[st]);
        recans(st + 1, a, s, nums);

        // Backtrack and make the recursive call excluding the current element
        a.pop_back();
        recans(st + 1, a, s, nums);
    }

    vector<vector<int>> subsets(vector<int>& nums) {
        set<vector<int>> s;
        vector<int> a;

        // Start the recursive process
        recans(0, a, s, nums);

        // Convert set to vector<vector<int>>
        return vector<vector<int>>(s.begin(), s.end());
    }
};

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