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, 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

    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