Introduction

In this blog post, we will explore how to solve the LeetCode problem "2215" 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 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where: answer[0] is a list of all distinct integers in nums1 which are not present in nums2. answer[1] is a list of all distinct integers in nums2 which are not present in nums1. Note that the integers in the lists may be returned in any order. Example 1: Input: nums1 = [1,2,3], nums2 = [2,4,6] Output: [[1,3],[4,6]] Explanation: For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3]. For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums1. Therefore, answer[1] = [4,6]. Example 2: Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2] Output: [[3],[]] Explanation: For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3]. Every integer in nums2 is present in nums1. Therefore, answer[1] = []. Constraints: 1 <= nums1.length, nums2.length <= 1000 -1000 <= nums1[i], nums2[i] <= 1000

Explanation

• Use sets for efficient lookup: Convert both input arrays into sets. Sets provide O(1) average time complexity for checking the existence of an element.
  • Iterate and compare: Iterate through each element of the first set and check if it exists in the second set. Add it to the result list if it doesn't. Repeat the process for the second set.

• Runtime Complexity: O(m + n), where m and n are the lengths of nums1 and nums2 respectively. Storage Complexity: O(m + n) in the worst case, to store the sets and the result lists.

Code

    def find_disappeared_numbers(nums1, nums2):
    """
    Finds the distinct integers in nums1 not present in nums2, and vice versa.

    Args:
        nums1: A list of integers.
        nums2: A list of integers.

    Returns:
        A list containing two lists:
        - The first list contains distinct integers in nums1 not present in nums2.
        - The second list contains distinct integers in nums2 not present in nums1.
    """

    set1 = set(nums1)
    set2 = set(nums2)

    result1 = []
    for num in set1:
        if num not in set2:
            result1.append(num)

    result2 = []
    for num in set2:
        if num not in set1:
            result2.append(num)

    return [result1, result2]