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] <= 10The challenge in #90 Subsets II is generating all possible subsets from an array that may contain duplicates, while ensuring the final result only includes unique subsets. A common strategy is to first sort the array, which groups duplicate elements together and makes it easier to avoid repeated subset generation.
The most effective approach uses backtracking. Starting from an empty subset, you recursively explore choices by including elements one by one. When iterating through the array, if the current element is the same as the previous one and the previous element was skipped at the same recursion level, you skip it to prevent duplicate subsets.
An alternative idea is using bit manipulation to generate all subset combinations, but additional checks or set structures are needed to filter duplicates. Backtracking with sorting is typically cleaner and more efficient.
Overall, the algorithm explores all subset combinations, resulting in exponential growth as the input size increases.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Backtracking with Sorting | O(n * 2^n) | O(n) |
| Bit Manipulation + Deduplication | O(n * 2^n) | O(2^n) |
take U forward
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.
1import java.util.ArrayList;
2import java.util.Arrays;
3import java.util.List;
4
5public
This Java solution sorts the array initially to help in skipping duplicates effectively. The backtracking method is implemented to recursively build subsets, ensuring duplicates are bypassed by checking for equality with the preceding element.
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.
1function
Watch expert explanations and walkthroughs
Practice problems asked by these companies to ace your technical interviews.
Explore More ProblemsJot down your thoughts, approach, and key learnings
Sorting places duplicate numbers next to each other. This allows the algorithm to easily detect and skip duplicates during the same recursion level, preventing identical subsets from being produced.
Yes, variations of the Subsets problem frequently appear in coding interviews at major tech companies. Interviewers use it to test understanding of recursion, backtracking, and handling duplicates in combinatorial problems.
The optimal approach is backtracking combined with sorting. Sorting groups duplicate elements together so the algorithm can skip repeated values during recursion, ensuring only unique subsets are generated.
A dynamic list or array is typically used to build subsets during backtracking. Recursion manages the exploration of combinations, while the result is stored in a list of lists.
JavaScript uses an iterative approach to handle subset creation with duplicates screening achieved by comprehensive sort operations. It processes existing subsets iteratively, precluding any repeat through index limitations tied to prior loops.