Problem Description

The task is to return all possible subsets (also known as the power set) of a given integer array nums, which may contain duplicates. The important constraint is that the solution set must not include duplicate subsets. These subsets can be returned in any order. A subset is a set containing elements that are all found in another set, which in this case is the array nums. The power set is the set of all subsets including the empty set and the set itself.

For example, if nums is [1,2,2], the possible unique subsets without considering order are:

- []
- [1]
- [2]
- [1,2]
- [2,2]
- [1,2,2]

The challenge here is to ensure that while generating the subsets, duplicates are not created.

Flowchart Walkthrough

Let's use the algorithm flowchart to analyze Leetcode problem 90, Subsets II, with the goal of determining the appropriate use of the backtracking pattern. Follow the Flowchart link for the decision map we'll navigate through:

Step-by-step Walkthrough:

1. Is it a graph?

   • Answer: No. The problem does not involve representing relationships between nodes and edges. Subsets generation is a combinatorial problem.

2. Need to solve for kth smallest/largest?

   • Answer: No. The problem is about generating all unique combinations, not finding a specific ordered element.

3. Involves Linked Lists?

   • Answer: No. The problem is based on array manipulation.

4. Does the problem have small constraints?

   • Answer: Yes. Although it involves generating subsets, which can exponentially grow, the input size itself is usually small enough to consider brute force or backtracking feasible.

5. Brute force / Backtracking?

   • Answer: Yes. To generate all possible unique subsets, considering the presence of duplicates and the need for uniqueness, backtracking is particularly suitable. Backtracking allows us to explore each possibility, backtrack on our choices, and prevent duplicate entries effectively.

Conclusion: The flowchart clearly guides us to the necessity of using a backtracking approach for solving the Subsets II problem effectively, further substantiated by the systematic exclusions of alternative strategies based on the specific problem's characteristics and requirements.

Intuition

The core idea to avoid duplicates in the subsets is to sort the array nums. Sorting brings identical elements next to each other, making it easier to avoid duplicates when constructing the subsets.

The solution uses Depth-First Search (DFS) to explore all the possible subsets. The process starts with an empty subset, and at each level, it includes the next number from the array into the current subset, then recursively continues to add the remaining numbers to form new subsets. This is done until all numbers have been considered. After each number is processed for a given subset, it is removed again (this operation is also known as backtracking), and the next number is tried.

To ensure no duplicated subsets are generated, there are two conditions:

1. Sort the Numbers: As mentioned, we start by sorting the array to bring duplicates together.

2. Skip over duplicates: While generating subsets, if the current element is the same as the previous element and the previous element was not included in the current subset being generated (checked by i != u), it is skipped to avoid creating a subset that has already been created.

This way, the dfs function systematically constructs all subsets, but avoids any repeats, ensuring that the power set it returns is free from duplicate subsets.

Learn more about Backtracking patterns.

Solution Approach

The implementation of the solution uses recursion to generate the subsets and backtracking to make sure all possible subsets are considered. Here's how the implementation works, with reference to the given Python code above:

1. Sort the Array: The first thing the solution does is to sort nums. This is crucial because it ensures that duplicates are adjacent and can be easily skipped when generating subsets.

   nums.sort()

2. Depth-First Search (DFS): The dfs function is then defined. DFS is a common algorithm for exploring all the possible configurations of a given problem. In this case, it's used to generate all possible subsets by making a series of choices (to include or not include an element in the current subset).

3. Recursive Exploration: Inside dfs, the current subset t is first included in the answer set ans. This represents the choice of not including any more elements in the current subset.

   ans.append(t[:])

4. Iterative Inclusion: The function then iterates through the remaining numbers starting from index u. If the number at the current index is the same as the one before it and it's not the first iteration of the loop (checked by i != u), it skips adding this number to avoid duplicates.

   for i in range(u, len(nums)):
       if i != u and nums[i] == nums[i - 1]:
           continue
       t.append(nums[i])
       dfs(i + 1, t)
       t.pop()

   • t.append(nums[i]) adds the current number to the subset.
   • A recursive call dfs(i + 1, t) is made to consider the next number in the array. The index i + 1 ensures that the function does not consider the current number again, thus moving forward in the array.
   • t.pop() is the backtracking step, which removes the last added number from the subset, allowing the function to consider other possibilities after returning from the recursive call.

5. Global Answer Set: An empty list ans is initialized outside the dfs function. As the subsets are created, they are added to ans. Since ans is defined outside the scope of the dfs function, it retains all the solutions across recursive calls.

   ans = []

6. Kick Start the DFS: The dfs function is initially called with the starting index 0 and an empty subset [] to start the exploration.

   dfs(0, [])

At the end of the execution, the global ans contains all the subsets without duplicates, which is then returned as the final result.

Example Walkthrough

Let's walk through an example to illustrate the solution approach using the array [2,1,2].

1. Sort the Array: First, sort the array to become [1,2,2]. Sorting ensures any duplicates are next to each other which helps later in the process to avoid duplicate subsets.

2. Depth-First Search (DFS): We define a dfs function that will be used to explore possible subsets.

3. Recursive Exploration: The exploration starts with an empty subset []. This subset is immediately added to the answer set ans because it is a valid subset (the empty set).

4. Iterative Inclusion: For the first call to dfs, we will look at each element starting with the first element of the sorted array.

   • Include the first element [1]. At this stage, dfs is recursively called to consider the next number.
   • At the next level, we have [1,2]. We include it and then the dfs explores further.
   • Now we have [1,2,2], which is also included.
   • After considering each of these, we backtrack to [1,2] and remove the last element to reconsider our options.
   • Since the next element is the same as the previous (as we just removed it), the loop skips adding [1,2] again because it would be a duplicate.

5. Backtracking: After the above steps, we backtrack to an empty subset [] and proceed to the next element:

   • The next element is [2]. We include it, and then dfs is called recursively.
   • At the next level, the function sees another 2. However, since the previous element is also 2, and we are not at the start of a new subset creation (checked by i != u), the algorithm takes care not to include [2,2] as this is already considered. This is where the duplicate gets skipped.

6. Global Answer Set: All the while, subsets like [], [1], [1,2], [1,2,2], and [2] are being added to the global ans list.

7. Complete Exploration: After the function explores all possibilities, the ans now contains, without duplicates:

- []
- [1]
- [2]
- [1,2]
- [1,2,2]

8. Return Results: Finally, the dfs has finished exploring, and ans with all the unique subsets is returned as the result.

This example demonstrates how the algorithm ensures all unique subsets of [2,1,2] are found by methodically exploring all options while skipping over duplicates due to sorting and the extra conditional skip in the recursive loop.

Solution Implementation

Time and Space Complexity

The given Python code defines a method subsetsWithDup that finds all possible subsets of a given set of numbers that might contain duplicates. To avoid duplicate subsets, the algorithm sorts the array and skips over duplicates during the depth-first search (DFS).

Time Complexity:

The time complexity of this algorithm can be analyzed based on the number of recursive calls and the operations performed on each call.

1. The function dfs is called recursively, at most, once for every element starting from the current index u. In the worst case, this means potentially calling dfs for each subset of nums.

2. Since nums has n elements, there are 2^n possible subsets including the empty subset.

3. However, because the list is sorted and the algorithm skips duplicate elements within the same recursive level, the number of recursive calls may be less than 2^n.

4. The append and pop methods of a list run in O(1) time.

Therefore, the time complexity of the algorithm is essentially bounded by the number of recursive calls and is O(2^n) in the worst case, when all elements are unique.

Space Complexity:

The space complexity is considered based on the space used by the recursion stack and the space needed to store the subsets (ans).

1. The maximum depth of the recursive call stack is n, which is the length of nums. Each recursive call uses space for local variables, which is O(n) space in the stack at most.

2. The list ans will eventually contain all subsets, and thus, it will have 2^n elements, considering each subset as an element. However, the total space taken by all subsets combined is also considerable as each of the 2^n subsets could have up to n elements. This effectively adds up to O(n * 2^n) space.

Considering these factors, the space complexity of the code can be defined as O(n * 2^n) because the space used by the algorithm is proportional to the number of subsets generated, and each subset at most can have n elements.

Learn more about how to find time and space complexity quickly using problem constraints.