# Solving Leetcode Interviews in Seconds with AI: Find the Maximum Divisibility Score


	# 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](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 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
	```python
	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
	```