Subsets II
<p>Given an integer array <code>nums</code> that may contain duplicates, 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 = [1,2,2]
<strong>Output:</strong> [[],[1],[1,2],[1,2,2],[2],[2,2]]
</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>
</ul>
Bit Manipulation; Array; Backtracking
function subsetsWithDup(nums: number[]): number[][] {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const t: number[] = [];
    const ans: number[][] = [];
    const dfs = (i: number): void => {
        if (i >= n) {
            ans.push([...t]);
            return;
        }
        t.push(nums[i]);
        dfs(i + 1);
        t.pop();
        while (i + 1 < n && nums[i] === nums[i + 1]) {
            i++;
        }
        dfs(i + 1);
    };
    dfs(0);
    return ans;
}
