Introduction

In this blog post, we will explore how to solve the LeetCode problem "2644" 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 two integer arrays nums and divisors. The divisibility score of divisors[i] is the number of indices j such that nums[j] is divisible by divisors[i]. Return the integer divisors[i] with the maximum divisibility score. If multiple integers have the maximum score, return the smallest one. Example 1: Input: nums = [2,9,15,50], divisors = [5,3,7,2] Output: 2 Explanation: The divisibility score of divisors[0] is 2 since nums[2] and nums[3] are divisible by 5. The divisibility score of divisors[1] is 2 since nums[1] and nums[2] are divisible by 3. The divisibility score of divisors[2] is 0 since none of the numbers in nums is divisible by 7. The divisibility score of divisors[3] is 2 since nums[0] and nums[3] are divisible by 2. As divisors[0], divisors[1], and divisors[3] have the same divisibility score, we return the smaller one which is divisors[3]. Example 2: Input: nums = [4,7,9,3,9], divisors = [5,2,3] Output: 3 Explanation: The divisibility score of divisors[0] is 0 since none of numbers in nums is divisible by 5. The divisibility score of divisors[1] is 1 since only nums[0] is divisible by 2. The divisibility score of divisors[2] is 3 since nums[2], nums[3] and nums[4] are divisible by 3. Example 3: Input: nums = [20,14,21,10], divisors = [10,16,20] Output: 10 Explanation: The divisibility score of divisors[0] is 2 since nums[0] and nums[3] are divisible by 10. The divisibility score of divisors[1] is 0 since none of the numbers in nums is divisible by 16. The divisibility score of divisors[2] is 1 since nums[0] is divisible by 20. Constraints: 1 <= nums.length, divisors.length <= 1000 1 <= nums[i], divisors[i] <= 109

Explanation

Here's the approach:

• Iterate through each divisor in the divisors array.
• For each divisor, iterate through the nums array and count how many numbers are divisible by the current divisor.

• Keep track of the divisor with the maximum divisibility score. If multiple divisors have the same maximum score, choose the smallest one.

• Runtime Complexity: O(n*m), where n is the length of nums and m is the length of divisors. Storage Complexity: O(1).

Code

    def maxDivScore(nums, divisors):
    max_score = -1
    result = -1

    for divisor in divisors:
        score = 0
        for num in nums:
            if num % divisor == 0:
                score += 1

        if score > max_score:
            max_score = score
            result = divisor
        elif score == max_score:
            result = min(result, divisor)

    return result