# Solving Leetcode Interviews in Seconds with AI: Intersection of Multiple Arrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2248" 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
	> Given a 2D integer array nums where nums[i] is a non-empty array of distinct positive integers, return the list of integers that are present in each array of nums sorted in ascending order.   Example 1:  Input: nums = [[3,1,2,4,5],[1,2,3,4],[3,4,5,6]] Output: [3,4] Explanation:  The only integers present in each of nums[0] = [3,1,2,4,5], nums[1] = [1,2,3,4], and nums[2] = [3,4,5,6] are 3 and 4, so we return [3,4]. Example 2:  Input: nums = [[1,2,3],[4,5,6]] Output: [] Explanation:  There does not exist any integer present both in nums[0] and nums[1], so we return an empty list [].    Constraints:  1 <= nums.length <= 1000 1 <= sum(nums[i].length) <= 1000 1 <= nums[i][j] <= 1000 All the values of nums[i] are unique.  

	# Explanation
	Here's the breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Use a hash table (dictionary in Python) to store the frequency of each number across all arrays.
    *   Iterate through the input `nums` and update the frequency count in the hash table for each number encountered.
    *   Filter the hash table to keep only the numbers whose frequency equals the number of arrays (meaning they are present in all arrays). Sort the result in ascending order.

*   **Complexity:**
    *   Runtime Complexity: O(N + M log M), where N is the total number of elements in all arrays and M is the number of common elements. The O(N) comes from iterating through the input arrays and updating the frequency in hashmap. The O(M log M) comes from sorting the common elements.
    *   Storage Complexity: O(N), where N is the number of distinct elements in the input array, which is for storing the frequency map.

	
	# Code
	```python
	def intersection(nums):
    """
    Finds the list of integers that are present in each array of nums sorted in ascending order.

    Args:
        nums: A 2D integer array where nums[i] is a non-empty array of distinct positive integers.

    Returns:
        A list of integers that are present in each array of nums sorted in ascending order.
    """

    counts = {}
    num_arrays = len(nums)

    for arr in nums:
        for num in arr:
            counts[num] = counts.get(num, 0) + 1

    result = [num for num, count in counts.items() if count == num_arrays]
    result.sort()
    return result
	```