3002. Maximum Size of a Set After Removals

MediumGreedyArrayHash Table
Leetcode Link

Problem Description

In this problem, you are given two arrays nums1 and nums2, each having an even length n. Your task is to remove half of the elements from each array (n / 2 from nums1 and n / 2 from nums2), and then combine the remaining elements into a set s. The goal is to maximize the size of the set s. Since sets do not allow duplicate values, any repeated values from nums1 and nums2 will only appear once in the set s. We're looking for the strategy that yields the largest possible set size after these removals and insertions are performed.


The key to solving this problem is understanding that if a number is unique to one array and not present in the other, it has to be included in the final set in order to maximize its size. Conversely, common elements shared between both arrays are already ensuring a unique entry in the set s, whether they are removed or not. Thus, the focus should be on the exclusive elements in each array first.

To achieve the maximum set size, you should:

  1. Convert each array to a set to remove duplicate values within the same array and to have efficient operations for following steps.
  2. Calculate the exclusive elements in each array using set difference operations (s1 - s2 and s2 - s1), and find out how many of these exclusive elements can be included in the set without exceeding the limit of n / 2 removals.
  3. The intersection of s1 and s2 gives the shared elements, which should also be included in the final set to maximize its size.
  4. The maximum possible size of the final set s will be the sum of the number of unique elements we can keep from s1, the number of unique elements we can keep from s2, and the number of shared elements between nums1 and nums2. This number, however, cannot exceed n because you are only allowed to keep n elements in total from both arrays combined after n / 2 removals from each.

The provided solution is direct and works efficiently by utilizing set operations in Python to find unique and shared elements and calculating the result accordingly.

Learn more about Greedy patterns.

Solution Approach

The solution approaches the problem by leveraging Python sets and basic mathematical computations to maximize the size of the combined set. The algorithm follows these steps:

  1. Convert Arrays to Sets: The first step is to convert both nums1 and nums2 arrays into sets s1 and s2. This operation removes any duplicates within individual arrays and allows for set operations to be performed in the next steps.

    1s1 = set(nums1)
    2s2 = set(nums2)
  2. Find Unique Elements: The solution calculates the number of unique elements in each set that are not present in the other set.

    1unique_to_s1 = s1 - s2
    2unique_to_s2 = s2 - s1

    This is done using the set difference operation (-). The length of these unique elements is then limited to n // 2 because that's the maximum number of elements that can be removed from each array.

    1a = min(len(unique_to_s1), n // 2)
    2b = min(len(unique_to_s2), n // 2)
  3. Compute Shared Elements: The solution determines the number of shared elements by computing the intersection of s1 and s2.

    1shared_elements = s1 & s2

    These shared elements contribute to the set's size because they would be unique in the combined set.

  4. Calculate Maximum Set Size: Finally, the maximum possible size of the set s is the sum of the unique elements that can be included from both s1 and s2, plus the number of shared elements. However, this cannot exceed n, which is the total number of elements that can exist in the set after removals.

    1return min(a + b + len(shared_elements), n)

The algorithm effectively uses set theory concepts to arrive at an efficient solution. Since operations on sets such as difference and intersection are typically much faster compared to list operations, and considering the limits on the number of elements that can be included after removals, this approach ensures that the resulting set is as large as possible according to the problem's constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

How would you design a stack which has a function min that returns the minimum element in the stack, in addition to push and pop? All push, pop, min should have running time O(1).

Example Walkthrough

Let's illustrate the solution approach with a small example using two arrays nums1 and nums2.

Assume nums1 = [1, 2, 2, 3] and nums2 = [2, 3, 4, 5], and they both have a length of 4. That means, we'll need to remove half of the elements from each array (2 elements from each), and then combine the remaining elements to maximize the size of the set s.

  1. Convert Arrays to Sets: By converting nums1 and nums2 into sets, we eliminate any duplicates within them and can perform set operations efficiently.

    1s1 = set([1, 2, 2, 3])   # s1 becomes set([1, 2, 3])
    2s2 = set([2, 3, 4, 5])   # s2 becomes set([2, 3, 4, 5])
  2. Find Unique Elements: We find the elements unique to each set by using set difference operations.

    1unique_to_s1 = s1 - s2    # unique_to_s1 becomes set([1]) as 1 is only in s1
    2unique_to_s2 = s2 - s1    # unique_to_s2 becomes set([4, 5]) as 4 and 5 are only in s2

    Then we limit the number of these unique elements to n // 2, which is 2 in this case.

    1a = min(len(unique_to_s1), 2)  # a is 1 because there is only 1 unique element in s1
    2b = min(len(unique_to_s2), 2)  # b is 2 because there are 2 unique elements in s2, within our limit
  3. Compute Shared Elements: We calculate the shared elements which is the intersection of the two sets.

    1shared_elements = s1 & s2   # shared_elements becomes set([2, 3])
  4. Calculate Maximum Set Size: The maximum possible size of the set s is computed by summing up the unique and shared elements.

    1max_set_size = min(a + b + len(shared_elements), 4)  # max_set_size is 4, which is the length of `nums1` and `nums2`

The resulting max_set_size is 4, which means we can keep all the three elements from s1 (as one of the elements '2' is common and only counted once) and two unique elements from s2 ('4' and '5') to form the set [1, 2, 3, 4, 5]. However, since we need to abide by the rule of removing 2 elements from each array, we would remove '2' and '3' from either nums1 or nums2 to respect the problem's constraints. Thus, the final size of set s is maximized within the given rules.

Solution Implementation

1class Solution:
2    def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:
3        # Convert lists into sets for efficient set operations
4        set_nums1 = set(nums1)
5        set_nums2 = set(nums2)
7        # Get the length of the first list
8        list_length = len(nums1)
10        # Calculate unique elements in nums1 not in nums2
11        # Limit the count to half of nums1's length because we aim for balanced numbers
12        unique_nums1 = min(len(set_nums1 - set_nums2), list_length // 2)
14        # Calculate unique elements in nums2 not in nums1
15        # Limit the count to half of nums1's length for the same reason as above
16        unique_nums2 = min(len(set_nums2 - set_nums1), list_length // 2)
18        # Count elements common to both sets
19        common_elements = len(set_nums1 & set_nums2)
21        # Maximize set size by combining the unique and common elements
22        # Ensure the combined size does not exceed the length of nums1
23        max_set_size = min(unique_nums1 + unique_nums2 + common_elements, list_length)
25        return max_set_size
1class Solution {
2    public int maximumSetSize(int[] nums1, int[] nums2) {
3        // Use HashSet to remove duplicates from both arrays
4        Set<Integer> setNums1 = new HashSet<>();
5        Set<Integer> setNums2 = new HashSet<>();
7        // Add all elements from nums1 to setNums1
8        for (int num : nums1) {
9            setNums1.add(num);
10        }
12        // Add all elements from nums2 to setNums2
13        for (int num : nums2) {
14            setNums2.add(num);
15        }
17        // Length of array nums1, given all arrays are same length
18        int length = nums1.length;
20        // Initialize counters for unique elements in setNums1 (a),
21        // in setNums2 (b), and common elements in both (c)
22        int uniqueNums1 = 0, uniqueNums2 = 0, commonElements = 0;
24        // Count unique elements in setNums1
25        for (int num : setNums1) {
26            if (!setNums2.contains(num)) {
27                ++uniqueNums1;
28            }
29        }
31        // Count unique elements in setNums2 and common elements
32        for (int num : setNums2) {
33            if (!setNums1.contains(num)) {
34                ++uniqueNums2;
35            } else {
36                ++commonElements;
37            }
38        }
40        // We can have at most n/2 unique elements from each set
41        uniqueNums1 = Math.min(uniqueNums1, length / 2);
42        uniqueNums2 = Math.min(uniqueNums2, length / 2);
44        // Maximum size of the set we can choose is sum of unique elements from both sets plus common elements.
45        // We ensure not to exceed the original length of the array.
46        return Math.min(uniqueNums1 + uniqueNums2 + commonElements, length);
47    }
1#include <vector>
2#include <unordered_set>
4class Solution {
6    int maximumSetSize(std::vector<int>& nums1, std::vector<int>& nums2) {
7        // Creating sets from the input vectors to eliminate duplicates and allow for efficient lookups.
8        std::unordered_set<int> uniqueNums1(nums1.begin(), nums1.end());
9        std::unordered_set<int> uniqueNums2(nums2.begin(), nums2.end());
11        // Store the size of the first vector.
12        int maxSize = nums1.size();
14        // Variables to keep count of unique elements in each set and common elements between the sets.
15        int uniqueInNums1 = 0, uniqueInNums2 = 0, commonElements = 0;
17        // Count elements unique to nums1's set.
18        for (int element : uniqueNums1) {
19            if (uniqueNums2.count(element) == 0) {
20                ++uniqueInNums1;
21            }
22        }
24        // Count elements unique to nums2's set and those that are common.
25        for (int element : uniqueNums2) {
26            if (uniqueNums1.count(element) == 0) {
27                ++uniqueInNums2;
28            } else {
29                ++commonElements;
30            }
31        }
33        // Ensure uniqueInNums1 count does not exceed half of maxSize.
34        uniqueInNums1 = std::min(uniqueInNums1, maxSize / 2);
35        // Ensure uniqueInNums2 count does not exceed half of maxSize.
36        uniqueInNums2 = std::min(uniqueInNums2, maxSize / 2);
38        // The maximum size is determined by the sum of unique and common elements but cannot exceed maxSize.
39        return std::min(uniqueInNums1 + uniqueInNums2 + commonElements, maxSize);
40    }
1function maximumSetSize(nums1: number[], nums2: number[]): number {
2    // Convert the input arrays to sets to eliminate duplicates
3    const setNums1: Set<number> = new Set(nums1);
4    const setNums2: Set<number> = new Set(nums2);
6    // Establish the length constraint for the subsets
7    const maxLength: number = nums1.length;
9    // Initialize counters for elements unique to each set and shared elements
10    let uniqueNums1: number = 0;
11    let uniqueNums2: number = 0;
12    let commonElements: number = 0;
14    // Count elements unique to the first set
15    for (const num of setNums1) {
16        if (!setNums2.has(num)) {
17            uniqueNums1++;
18        }
19    }
21    // Count elements unique to the second set and shared elements
22    for (const num of setNums2) {
23        if (!setNums1.has(num)) {
24            uniqueNums2++;
25        } else {
26            commonElements++;
27        }
28    }
30    // The number of unique elements we can take from each set is limited by half the length of nums1
31    uniqueNums1 = Math.min(uniqueNums1, maxLength >> 1);
32    uniqueNums2 = Math.min(uniqueNums2, maxLength >> 1);
34    // The maximum size of the resulting set is the sum of unique and common elements,
35    // but it can't exceed the length of nums1
36    return Math.min(uniqueNums1 + uniqueNums2 + commonElements, maxLength);

Time and Space Complexity

Time Complexity

The time complexity of the maximumSetSize function mainly stems from a few operations:

  1. Converting nums1 to a set s1 and nums2 to a set s2 requires O(N) time each, where N is the number of elements in the respective lists.
  2. Finding the difference between sets (s1 - s2 and s2 - s1) and their intersection (s1 & s2) have a time complexity of O(N) each, under the assumption that the elements in the sets are hashed properly.
  3. The min function calls are constant time operations and do not significantly contribute to the overall time complexity.

Hence, the overall time complexity is O(N), considering all N elements would have to be processed to convert them into sets and perform set operations.

Space Complexity

The space complexity is dominated by the additional sets we create:

  1. s1 and s2 both require O(N) space, where N is the length of nums1 and nums2 respectively, as in the worst-case scenario they could be all unique.
  2. Temporary space for set operations is also O(N) in the worst case when all elements are unique.

Therefore, the overall space complexity is also O(N), where N is the size of the larger input list because we need space to store unique elements of each input list.

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

Fast Track Your Learning with Our Quick Skills Quiz:

Which one best describes the time complexity of the following code?

1int factorial(int n) {
2  if (n < 0) {
3    return -1;
4  } else if (n == 0) {
5    return 1;
6  } else {
7    return n * factorial(n - 1);
8  }

Recommended Readings

Got a question? Ask the Monster Assistant anything you don't understand.

Still not clear? Ask in the Forum,  Discord or Submit the part you don't understand to our editors.

Tired of the LeetCode Grind?

Our structured approach teaches you the patterns behind problems, so you can confidently solve any challenge. Get started now to land your dream tech job.

Get Started