<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:
    string stringify(vector<int> v) {
        string s;
        for (int i : v) {
            s += to_string(i) + ' ';
        }
        return s;
    }

    vector<vector<int>> res;
    void backtrack(vector<int> options, vector<int> comb, int start, int n) {
        res.push_back(comb);
        cout << "comb: " << stringify(comb) << "; start: " << start << "; options: " << stringify(options) << '\n';
        for (int i = start; i < n; i++) {
            int val = options[i];
            if (val == -11) continue;
            comb.push_back(val);
            // options.erase(options.begin() + i);
            options[i] = -11;
            backtrack(options, comb, i+1, n);
            // options.insert(options.begin() + i, val);
            options[i] = val;
            comb.pop_back();
        }
    }

    vector<vector<int>> subsets(vector<int>& nums) {
        vector<int> v;
        backtrack(nums, v, 0, nums.size());
        return res;
    }
};
12600