How do you solve Subsets II on LeetCode?

Subsets II (LeetCode #90) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n) space complexity. Key insight: Sorting the array helps in efficiently skipping duplicates.

What is the brute force solution for Subsets II?

The brute force approach for Subsets II is Brute Force, with O(n * 2^n) time complexity and O(2^n) space complexity. The brute-force approach generates all possible subsets of the input array, including duplicates. It does this by checking every combination of elements, which can be thought of as flipping a coin for each element: include it or not. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Subsets II?

Subsets II uses the following concepts: Array, Backtracking, Bit Manipulation. The recommended patterns to study are: Backtracking, Sorting.

What are the interview tips for Subsets II?

Explain your thought process clearly while coding. Discuss edge cases, such as an empty array or all elements being the same. Practice writing clean and efficient code with proper variable names.

What are common mistakes when solving Subsets II?

Not sorting the array before processing, leading to duplicates in the result. Failing to handle the base case in recursion properly.

#90

Subsets II

Medium

ArrayBacktrackingBit ManipulationBacktrackingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * 2^n)	O(n * 2^n)
Space	O(2^n)	O(n)

💡

Intuition

Time O(n * 2^n)Space O(n)

The optimal solution uses backtracking to build subsets while ensuring that duplicates are avoided by skipping over duplicate elements. This is more efficient than generating all subsets and filtering them afterward.

⚙️

Algorithm

3 steps

1Step 1: Sort the input array to group duplicates together.
2Step 2: Use a backtracking function to build subsets, adding elements only if they are not duplicates of the previous element at the same level.
3Step 3: Add each valid subset to the result list.

solution.py13 lines

1def subsetsWithDup(nums):
2    nums.sort()
3    result = []
4    def backtrack(start, path):
5        result.append(path[:])
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    backtrack(0, [])
13    return result

ℹ

Complexity note: The time complexity remains O(n * 2^n) due to the nature of generating subsets, but the space complexity is O(n) for the recursion stack and the result storage.

1Sorting the array helps in efficiently skipping duplicates.
2Backtracking allows us to explore all combinations without generating all subsets upfront.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.