Introduction

In this blog post, we will explore how to solve the LeetCode problem "350" 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

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order. Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9] Explanation: [9,4] is also accepted. Constraints: 1 <= nums1.length, nums2.length <= 1000 0 <= nums1[i], nums2[i] <= 1000 Follow up: What if the given array is already sorted? How would you optimize your algorithm? What if nums1's size is small compared to nums2's size? Which algorithm is better? What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

Explanation

Here's the breakdown of the solution:

• Frequency Counting: Use a hash map (dictionary in Python) to store the frequency of each element in the smaller array (nums1). This allows for efficient lookups.
• Iterate and Match: Iterate through the larger array (nums2). If an element exists in the frequency map and its count is greater than 0, add it to the result and decrement its count in the frequency map.

• Handle Sorted Input (Optimization): If the arrays are sorted, a two-pointer approach is optimal. Maintain pointers for both arrays and advance them based on comparisons.

• Runtime Complexity: O(m + n), where m and n are the lengths of the input arrays. Storage Complexity: O(min(m, n)) in the average case (due to the hash map). O(1) if the array are sorted and we apply the two pointer approach.

Code

    def intersect(nums1, nums2):
    """
    Finds the intersection of two integer arrays, preserving element counts.

    Args:
        nums1: The first integer array.
        nums2: The second integer array.

    Returns:
        A list containing the intersection of the two arrays, with elements
        appearing as many times as they appear in both arrays.
    """

    if len(nums1) > len(nums2):
        return intersect(nums2, nums1)  # Ensure nums1 is the smaller array

    freq_map = {}
    for num in nums1:
        freq_map[num] = freq_map.get(num, 0) + 1

    result = []
    for num in nums2:
        if num in freq_map and freq_map[num] > 0:
            result.append(num)
            freq_map[num] -= 1

    return result