90. Subsets II
Description
Given an integer array nums
that may contain duplicates, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Example 1:
Input: nums = [1,2,2] Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]
Example 2:
Input: nums = [0] Output: [[],[0]]
Constraints:
1 <= nums.length <= 10
-10 <= nums[i] <= 10
Solutions
Solution 1: Sorting + DFS
We can first sort the array $nums$ to facilitate deduplication.
Then, we design a function $dfs(i)$, which represents searching for subsets starting from the $i$-th element. The execution logic of the function $dfs(i)$ is as follows:
If $i \geq n$, it means that all elements have been searched, and the current subset is added to the answer array, and the recursion ends.
If $i < n$, add the $i$-th element to the subset, execute $dfs(i + 1)$, and then remove the $i$-th element from the subset. Next, we judge whether the $i$-th element is the same as the next element. If it is the same, we loop to skip this element until we find the first element that is different from the $i$-th element, and execute $dfs(i + 1)$.
Finally, we only need to call $dfs(0)$ and return the answer array.
The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.
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Solution 2: Sorting + Binary Enumeration
Similar to Solution 1, we first sort the array $nums$ to facilitate deduplication.
Next, we enumerate a binary number $mask$ in the range of $[0, 2^n)$, where the binary representation of $mask$ is an $n$-bit bit string. If the $i$-th bit of $mask$ is $1$, it means to select $nums[i]$, and $0$ means not to select $nums[i]$. Note that if the $i - 1$ bit of $mask$ is $0$, and $nums[i] = nums[i - 1]$, it means that in the current enumerated scheme, the $i$-th element and the $i - 1$-th element are the same. To avoid repetition, we skip this situation. Otherwise, we add the subset corresponding to $mask$ to the answer array.
After the enumeration ends, we return the answer array.
The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.
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