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, 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

    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)