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Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set). The solution set must not contain duplicate subsets. Return the solution in any order.

Example:

Input: nums = [1,2,2]
Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]

Clarifying Questions

1. Q: Can the input list nums be empty?
   • A: Yes, if the input list is empty, the output should be [[]].
2. Q: Are the elements of the list always integers?
   • A: Yes, the problem statement specifies a collection of integers.
3. Q: Is there any constraint on the size of the input list?
   • A: There is no explicit constraint mentioned, but typical list sizes in coding problems can range up to hundreds or thousands. LeetCode usually specifies constraints in the full problem description.

Strategy

1. Sort the Input List:
   • Sorting helps to easily skip duplicates when generating subsets.
2. Backtracking Approach:
   • Use a backtracking algorithm to explore all possible subsets.
   • At each step in the recursion, choose to either include the current element or not.
   • To avoid duplicates, skip elements that are the same as the previous element.
   • Maintain a list to collect the current subset and add it to the final result.
3. Avoid Duplicates:
   • By sorting the list initially, we can easily compare the current element with the previous one and skip processing if they are the same.

Code

Here’s the Python code implementing the above strategy:

def subsetsWithDup(nums):
    def backtrack(start, path):
        # Add the current subset (path) to the result
        result.append(path[:])
        for i in range(start, len(nums)):
            # If current element is same as previous, skip it to avoid duplicates
            if i > start and nums[i] == nums[i - 1]:
                continue
            # Include nums[i] in the current subset
            path.append(nums[i])
            # Move onto the next element
            backtrack(i + 1, path)
            # Exclude nums[i] from the current subset (backtrack)
            path.pop()

    nums.sort()
    result = []
    backtrack(0, [])
    return result

# Example Usage
nums = [1, 2, 2]
print(subsetsWithDup(nums))  # Output: [[], [1], [1, 2], [1, 2, 2], [2], [2, 2]]

Time Complexity

• Sorting: Sorting the list initially takes (O(n \log n)), where (n) is the length of nums.
• Generating Subsets: In the worst case, there are (2^n) possible subsets.
  • Since for each subset, the backtracking function does work proportional to its size, generating all subsets takes (O(n \cdot 2^n)) time.
  • This is because each subset takes linear time to be processed at each step, and there are (2^n) subsets.
• Overall Complexity: (O(n \log n) + O(n \cdot 2^n) = O(n \cdot 2^n))

Space Complexity

• Auxiliary Space: The auxiliary space used by the recursion stack is (O(n)).
• Result Storage: The result list potentially contains (2^n) subsets, each of which might be up to length (n).
  • This results in (O(n \cdot 2^n)) space for storing all subsets.

Thus, the overall space complexity is also (O(n \cdot 2^n)).

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