Permutations II
<p>Given a collection of numbers, <code>nums</code>,&nbsp;that might contain duplicates, return <em>all possible unique permutations <strong>in any order</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,2]
<strong>Output:</strong>
[[1,1,2],
 [1,2,1],
 [2,1,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 8</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
</ul>
Array; Backtracking; Sorting
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        ranges::sort(nums);
        int n = nums.size();
        vector<vector<int>> ans;
        vector<int> t(n);
        vector<bool> vis(n);
        auto dfs = [&](this auto&& dfs, int i) {
            if (i == n) {
                ans.emplace_back(t);
                return;
            }
            for (int j = 0; j < n; ++j) {
                if (vis[j] || (j && nums[j] == nums[j - 1] && !vis[j - 1])) {
                    continue;
                }
                t[i] = nums[j];
                vis[j] = true;
                dfs(i + 1);
                vis[j] = false;
            }
        };
        dfs(0);
        return ans;
    }
};
