# Solving Leetcode Interviews in Seconds with AI: Maximum Size of a Set After Removals


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3002" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Identify Common Elements:** Find the elements that exist in both `nums1` and `nums2`. These are the elements we potentially want to minimize removing from either array, since removing them reduces the final set size.

*   **Prioritize Uncommon Elements:** Count the occurrences of each element in each array. For elements that are *not* in the intersection (common elements), we *must* keep as many as possible after removing `n/2` elements from each array.

*   **Greedy Removal Strategy:** If `k` is n/2: For common elements, if the count in both arrays is greater than k, we remove enough elements from both arrays to make it equal to k for both. If the total elements in set is less than n, we keep all uncommon elements after removing the common elements from either array.

*   **Time Complexity: O(n), Storage Complexity: O(n)**
    where n is the length of the input arrays.

	
	# Code
	```python
	def maximumSetSize(nums1, nums2):
    n = len(nums1)
    k = n // 2

    set1 = set(nums1)
    set2 = set(nums2)
    common = set1.intersection(set2)

    count1 = {}
    count2 = {}

    for num in nums1:
        count1[num] = count1.get(num, 0) + 1
    for num in nums2:
        count2[num] = count2.get(num, 0) + 1

    remove1 = 0
    remove2 = 0

    for num in common:
        take1 = min(k, count1[num])
        take2 = min(k, count2[num])
        count1[num] = take1
        count2[num] = take2

        remove1 += take1
        remove2 += take2

    keep1 = k
    keep2 = k
    result = 0

    for num in set1:
        if num not in common:
            result += 1

    for num in set2:
        if num not in common and num not in set1:
            result += 1

    final_set = set()
    for num in nums1:
        if num in count1 and count1[num] > 0:
            final_set.add(num)
            count1[num] -=1
    for num in nums2:
        if num in count2 and count2[num] > 0:
            final_set.add(num)
            count2[num] -=1
    return len(final_set)
	```