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To solve this problem, we need to perform four common set operations on two input lists: union, intersection, difference (first list minus second list), and symmetric difference. These operations are efficiently handled using Python's set data structure, which inherently supports these operations and automatically handles duplicates.

Approach

1. Convert Lists to Sets: First, we convert both input lists into sets. This allows us to leverage Python's built-in set operations and eliminate duplicates.
2. Perform Set Operations:
   • Union: Combines elements from both sets (elements present in either set).
   • Intersection: Finds elements common to both sets.
   • Difference: Finds elements present in the first set but not in the second set.
   • Symmetric Difference: Finds elements present in either set but not in both.
3. Return Results: The results of these operations are returned as a tuple of four sets.

Solution Code

def list_operations(list1, list2):
    set1 = set(list1)
    set2 = set(list2)
    return (
        set1.union(set2),
        set1.intersection(set2),
        set1.difference(set2),
        set1.symmetric_difference(set2)
    )

Explanation

• Union: set1.union(set2) (or set1 | set2) returns a set containing all elements from both set1 and set2.
• Intersection: set1.intersection(set2) (or set1 & set2) returns a set containing elements that are in both set1 and set2.
• Difference: set1.difference(set2) (or set1 - set2) returns a set containing elements that are in set1 but not in set2.
• Symmetric Difference: set1.symmetric_difference(set2) (or set1 ^ set2) returns a set containing elements that are in either set1 or set2 but not in both.

This approach is efficient with a time complexity of O(n + m) where n and m are the lengths of the input lists, as converting a list to a set and performing set operations are linear time operations. The results are unordered (since sets are unordered) but contain all the correct elements as per the required operations.

Example Usage:

# Test case
list_a = [1, 2, 3, 3]
list_b = [3, 4, 5]
result = list_operations(list_a, list_b)
print(result)  
# Output: ({1, 2, 3, 4, 5}, {3}, {1, 2}, {1, 2, 4, 5})

This solution correctly handles duplicates in the input lists and provides the expected results for all four operations. The output is a tuple of sets, which is the standard way to represent these operations in Python. If you need the results as lists, you can convert each set to a list (e.g., list(set1.union(set2))), though the order of elements in the list will be arbitrary. However, the problem statement does not specify the output type (list vs set), so using sets is appropriate here.

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