How do you solve Maximum Size of a Set After Removals on LeetCode?

Maximum Size of a Set After Removals (LeetCode #3002) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Maximizing unique elements requires careful selection of which elements to keep.

What is the time complexity of Maximum Size of a Set After Removals?

The optimal solution for Maximum Size of a Set After Removals runs in O(n) time and uses O(n) space. The complexity is O(n) because we only pass through the arrays a couple of times to count elements and create a set of unique values.

What is the brute force solution for Maximum Size of a Set After Removals?

The brute force approach for Maximum Size of a Set After Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of elements to keep in both arrays. This allows us to check every possible way to remove elements and find the maximum unique elements in the resulting set. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Size of a Set After Removals?

Maximum Size of a Set After Removals uses the following concepts: Array, Hash Table, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Size of a Set After Removals?

Clearly explain your thought process while solving the problem. Consider edge cases, such as when both arrays have the same elements. Practice similar problems to become familiar with greedy strategies.

What are common mistakes when solving Maximum Size of a Set After Removals?

Not considering the frequency of elements when deciding which to keep. Failing to account for the total number of elements that can be kept from both arrays.

#3002

Maximum Size of a Set After Removals

Medium

ArrayHash TableGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a greedy strategy to maximize the unique elements in the resulting set by counting the occurrences of elements in both arrays and strategically choosing which elements to keep.

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Algorithm

3 steps

1Step 1: Count the frequency of each element in both nums1 and nums2.
2Step 2: Create a set of unique elements from both arrays.
3Step 3: Calculate the maximum possible size of the set by keeping unique elements while ensuring we only keep n/2 elements from each array.

solution.py9 lines

1# Full working Python code
2from collections import Counter
3
4def maxSetSize(nums1, nums2):
5    count1 = Counter(nums1)
6    count2 = Counter(nums2)
7    unique_elements = set(nums1) | set(nums2)
8    max_size = min(len(unique_elements), len(nums1) // 2 + len(nums2) // 2)
9    return max_size

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Complexity note: The complexity is O(n) because we only pass through the arrays a couple of times to count elements and create a set of unique values.

1Maximizing unique elements requires careful selection of which elements to keep.
2Using frequency counts helps in determining the best strategy for removals.

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