# Solving Leetcode Interviews in Seconds with AI: Find the Number of Good Pairs I


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3162" 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 2 integer arrays nums1 and nums2 of lengths n and m respectively. You are also given a positive integer k. A pair (i, j) is called good if nums1[i] is divisible by nums2[j] * k (0 <= i <= n - 1, 0 <= j <= m - 1). Return the total number of good pairs.   Example 1:  Input: nums1 = [1,3,4], nums2 = [1,3,4], k = 1 Output: 5 Explanation: The 5 good pairs are (0, 0), (1, 0), (1, 1), (2, 0), and (2, 2). Example 2:  Input: nums1 = [1,2,4,12], nums2 = [2,4], k = 3 Output: 2 Explanation: The 2 good pairs are (3, 0) and (3, 1).    Constraints:  1 <= n, m <= 50 1 <= nums1[i], nums2[j] <= 50 1 <= k <= 50  

	# Explanation
	Here's an efficient approach to solve the problem:

*   **Precompute Divisibility:** Iterate through `nums1` and `nums2`. For each pair `(nums1[i], nums2[j])`, check if `nums1[i]` is divisible by `nums2[j] * k`.
*   **Count Good Pairs:** Maintain a counter and increment it for each good pair found.
*   **Direct Calculation:**  Implement the divisibility check directly within the nested loops for efficiency.

*   **Runtime Complexity:** O(n\*m), where n and m are the lengths of `nums1` and `nums2` respectively. **Storage Complexity:** O(1).

	
	# Code
	```python
	def count_good_pairs(nums1, nums2, k):
    """
    Counts the number of good pairs (i, j) where nums1[i] is divisible by nums2[j] * k.

    Args:
        nums1: The first list of integers.
        nums2: The second list of integers.
        k: The positive integer.

    Returns:
        The total number of good pairs.
    """

    n = len(nums1)
    m = len(nums2)
    count = 0

    for i in range(n):
        for j in range(m):
            if nums1[i] % (nums2[j] * k) == 0:
                count += 1

    return count
	```