Maximum Size of a Set After Removals

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ArraysGreedy Algorithms

You are given two 0-indexed integer arrays nums1 and nums2 of even length n.

You must remove n / 2 elements from nums1 and n / 2 elements from nums2. After the removals, you insert the remaining elements of nums1 and nums2 into a set s.

Return the maximum possible size of the set s.

Example 1:

Input: nums1 = [1,2,1,2], nums2 = [1,1,1,1]
Output: 2
Explanation: We remove two occurences of 1 from nums1 and nums2. After the removals, the arrays become equal to nums1 = [2,2] and nums2 = [1,1]. Therefore, s = {1,2}.
It can be shown that 2 is the maximum possible size of the set s after the removals.

Example 2:

Input: nums1 = [1,2,3,4,5,6], nums2 = [2,3,2,3,2,3]
Output: 5
Explanation: We remove 2, 3, and 6 from nums1, as well as 2 and two occurrences of 3 from nums2. After the removals, the arrays become equal to nums1 = [1,4,5] and nums2 = [2,3,2]. Therefore, s = {1,2,3,4,5}.
It can be shown that 5 is the maximum possible size of the set s after the removals.

Example 3:

Input: nums1 = [1,1,2,2,3,3], nums2 = [4,4,5,5,6,6]
Output: 6
Explanation: We remove 1, 2, and 3 from nums1, as well as 4, 5, and 6 from nums2. After the removals, the arrays become equal to nums1 = [1,2,3] and nums2 = [4,5,6]. Therefore, s = {1,2,3,4,5,6}.
It can be shown that 6 is the maximum possible size of the set s after the removals.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 2 * 10⁴
n is even.
1 <= nums1[i], nums2[i] <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the arrays, n?
Can the arrays contain negative numbers, zeros, or any other non-positive values?
Are there any duplicate numbers within either nums1 or nums2, and how should those duplicates be handled when forming set S?
If n is odd, should I round n/2 up or down when determining the number of elements to remove?
Is the order of elements in nums1 and nums2 significant, or can I freely reorder them during the removal process?

Brute Force Solution

Approach

The brute force approach explores every possible way to remove elements from the two sets until we reach a balanced state where the sets are equal in size and represent the largest possible set. We consider all combinations of removals to find the optimal solution. It's like trying every possible combination of decisions to see what leads to the best outcome.

Here's how the algorithm would work step-by-step:

Consider taking out zero items, one item, two items, and so on, from the first set.
For each of those choices, consider all the possible ways to take out items from the second set.
After removing items from both sets in all the ways possible, find out how many unique items are left overall when you combine the remaining items.
Remember that we are allowed to combine them only when the remaining items in each set are equal in amount.
Keep track of the largest total number of unique items you have found from all the removal combinations that end up in equal sized sets.
The largest size you kept track of is your answer.

Code Implementation

def max_set_size_after_removals_brute_force(original_collection):
    maximum_set_size = 0

    # Iterate through all possible subsets for removal
    for i in range(1 << len(original_collection)):
        subset_to_remove = []
        for j in range(len(original_collection)):
            if (i >> j) & 1:
                subset_to_remove.append(original_collection[j])

        # Create a new list representing the collection after removal
        current_collection = original_collection[:]
        for element in subset_to_remove:
            current_collection.remove(element)

        # Check if the remaining elements form a set
        if len(current_collection) == len(set(current_collection)):
            # Track the maximum size found so far
            if len(current_collection) > maximum_set_size:
                maximum_set_size = len(current_collection)

    # Checking for an empty set on input
    if not original_collection:
        return 0
    # Return the maximum set size found after all removals
    return maximum_set_size

Big(O) Analysis

Time Complexity

O(2^(2n)) – The brute force approach explores all possible subsets from both sets. If each set has n elements, then there are 2^n possible subsets for each set. Since we iterate through all subsets of both sets independently, and for each pair of subsets check if their sizes are equal and calculate the size of the union of the remaining elements, the total number of operations is proportional to 2^n * 2^n, which is 2^(2n). Therefore, the time complexity is O(2^(2n)).

Space Complexity

O(2^N) – The brute force approach described explores all possible subsets of both input sets to find the optimal solution. This inherently involves considering every combination of removals from the first set and then from the second set. Because there are 2^N possible subsets for a set of size N, the space complexity stems from the implicit call stack created by the recursive exploration of these subsets. While the explanation does not explicitly mention recursion or data structures used to store these subsets directly, it implies an exhaustive exploration of the possibilities, which often manifests as a recursive algorithm with a call stack growing proportionally to the number of possibilities, or 2^N in the worst case where N represents the size of the larger of the two input sets.

Optimal Solution

Approach

The key is to understand that we can maximize the set size by strategically removing elements from both input sets to balance their sizes. The optimal approach focuses on eliminating duplicates first and then balancing the unique elements to achieve the largest possible set.

Here's how the algorithm would work step-by-step:

First, find all the elements that appear in both input sets. These are duplicates that count against the final set size.
Remove all of these duplicate elements, keeping track of how many you removed from each set.
Now, look at the sizes of the two sets. One set might be larger than the other.
If one set is larger, figure out how many elements you need to remove from it to make both sets the same size (up to the target set size).
Remove the needed number of elements from the larger set, prioritizing removing elements that have not already been removed.
The maximum size of the final set is the size of one of the sets after the removals. Since we aim to maximize the set size, we need to ensure the final size isn't greater than half of the total elements of the two given sets, as the final set can contain at most the number of unique elements in both input sets. To ensure the valid final size, we return the minimum value between the calculated maximum set size and the half of the total elements from the two input sets.

Code Implementation

def maximum_set_size_after_removals(nums1, nums2):
    set1 = set(nums1)
    set2 = set(nums2)

    unique_count1 = len(set1)
    unique_count2 = len(set2)
    total_length = len(nums1) 

    if unique_count1 + unique_count2 <= total_length:
        return unique_count1 + unique_count2

    intersection_size = len(set1.intersection(set2))
    
    # Handle the case where one set is much larger than allowed.
    if unique_count1 > total_length // 2:
        removal_needed = unique_count1 - total_length // 2

        # Prioritize removing elements that exist in both sets.
        elements_to_remove = min(removal_needed, intersection_size)
        unique_count1 -= elements_to_remove
        intersection_size -= elements_to_remove

        unique_count1 -= (removal_needed - elements_to_remove)

    if unique_count2 > total_length // 2:
        removal_needed = unique_count2 - total_length // 2

        elements_to_remove = min(removal_needed, intersection_size)
        unique_count2 -= elements_to_remove
        intersection_size -= elements_to_remove

        unique_count2 -= (removal_needed - elements_to_remove)

    # Ensure that the combined size does not exceed the total length.
    return min(unique_count1 + unique_count2, total_length)

Big(O) Analysis

Time Complexity

O(n) – Finding common elements between the two sets, removing duplicates and balancing the set sizes involve iterating through the elements of both sets. These operations such as finding the intersection of sets and removing elements occur in linear time relative to the size 'n' of the combined input sets. As such, the overall time complexity is dominated by these linear operations, resulting in a time complexity of O(n).

Space Complexity

O(N) – The algorithm uses hash sets (or similar data structures) to store the elements of the input sets and the intersection between them. In the worst case, where all elements in both input arrays are unique, the hash sets will store close to all elements from the two sets. Thus the space required is proportional to the total number of elements across both sets which we can define as N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input arrays (nums1 and nums2)	Return 0 as no elements can be added to the set.
Arrays with size n=2 (minimum non-trivial case)	The solution should correctly handle removing 1 element from each array and adding the remaining elements (at most 2) to the set.
Large input arrays (n approaching memory limits)	The solution should use memory efficiently, considering the space complexity of the set and any intermediate data structures.
Arrays containing duplicate values within nums1 or nums2	The set data structure inherently handles duplicates, ensuring only unique elements are added to the set.
Arrays containing negative numbers or zeros	The solution should work correctly with negative numbers and zeros, as the problem statement doesn't impose any restrictions on the range of values.
Arrays where nums1 and nums2 have significant overlap (many common elements)	The set will eliminate duplicates in this scenario, resulting in a smaller set size than if the arrays had disjoint elements.
Integer overflow when calculating n/2 for extremely large n	Use integer division which truncates as desired, and verify no other calculations could overflow.
Arrays with all identical elements in nums1 or nums2	The removal process and the set operation will handle this correctly, but could lead to performance differences depending on the specific algorithm used.

Empty input arrays (nums1 and nums2)

How to Handle:

Return 0 as no elements can be added to the set.

Arrays with size n=2 (minimum non-trivial case)

How to Handle:

The solution should correctly handle removing 1 element from each array and adding the remaining elements (at most 2) to the set.

Large input arrays (n approaching memory limits)

How to Handle:

The solution should use memory efficiently, considering the space complexity of the set and any intermediate data structures.

Arrays containing duplicate values within nums1 or nums2

How to Handle:

The set data structure inherently handles duplicates, ensuring only unique elements are added to the set.

Arrays containing negative numbers or zeros

How to Handle:

The solution should work correctly with negative numbers and zeros, as the problem statement doesn't impose any restrictions on the range of values.

Arrays where nums1 and nums2 have significant overlap (many common elements)

How to Handle:

The set will eliminate duplicates in this scenario, resulting in a smaller set size than if the arrays had disjoint elements.

Integer overflow when calculating n/2 for extremely large n

How to Handle:

Use integer division which truncates as desired, and verify no other calculations could overflow.

Arrays with all identical elements in nums1 or nums2

How to Handle:

The removal process and the set operation will handle this correctly, but could lead to performance differences depending on the specific algorithm used.