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
impl Solution {
    pub fn subsets_with_dup(nums: Vec<i32>) -> Vec<Vec<i32>> {
        let mut nums = nums;
        nums.sort();
        let mut ans = Vec::new();
        let mut t = Vec::new();

        fn dfs(i: usize, nums: &Vec<i32>, t: &mut Vec<i32>, ans: &mut Vec<Vec<i32>>) {
            if i >= nums.len() {
                ans.push(t.clone());
                return;
            }
            t.push(nums[i]);
            dfs(i + 1, nums, t, ans);
            t.pop();
            let mut i = i;
            while i + 1 < nums.len() && nums[i + 1] == nums[i] {
                i += 1;
            }
            dfs(i + 1, nums, t, ans);
        }

        dfs(0, &nums, &mut t, &mut ans);
        ans
    }
}
