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In this approach, the problem of subsets with duplicates is solved by first sorting the array. This assists in easily identifying and omitting duplicates. The backtracking method is utilized to construct all possible subsets, ensuring no duplicate subsets are generated by skipping duplicate numbers during recursive backtracking.
The time complexity of this solution is O(2^n * n) due to the number of subsets (2^n) and time taken to convert each subset to the result. The space complexity is O(n), where n is the length of the array, due to the space required for the subset building.
1from typing import List
2
3def subsetsWithDup(nums: List[int]) -> List[List[int]]:
4 def backtrack(start: int, path: List[int]):
5 result.append(path.copy())
6 for i in range(start, len(nums)):
7 if i > start and nums[i] == nums[i - 1]:
8 continue
9 path.append(nums[i])
10 backtrack(i + 1, path)
11 path.pop()
12
13 nums.sort()
14 result = []
15 backtrack(0, [])
16 return result
17
18# Example Usage
19d = subsetsWithDup([1, 2, 2])
20for subset in d:
21 print(subset)This Python solution also uses sorting followed by a recursive backtracking method to eliminate duplicate subsets. It accomplishes this by checking and skipping duplicates.
This approach iteratively constructs the power set by extending previously generated sets with each new number. During the process, duplicates are skipped by comparing the current number with the previous one. This ensures subsets generated in each iteration are unique.
The time complexity is O(2^n * n); space is O(2^n * n) as it iteratively builds the power set using subsets directly.
1import
In this Java solution, subsets are iteratively expanded. The sorting makes managing the duplicates straightforward. Subset collections dynamically grow, maintaining control over redundancy by observing and limiting duplicate origins.