You are given two 0-indexed 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 there is more than one integer with the maximum score, return the minimum of them.

Example 1:

Input: nums = [4,7,9,3,9], divisors = [5,2,3]
Output: 3
Explanation: The divisibility score for every element in divisors is:
The divisibility score of divisors is 0 since no number in nums is divisible by 5.
The divisibility score of divisors is 1 since nums is divisible by 2.
The divisibility score of divisors is 3 since nums, nums, and nums are divisible by 3.
Since divisors has the maximum divisibility score, we return it.


Example 2:

Input: nums = [20,14,21,10], divisors = [5,7,5]
Output: 5
Explanation: The divisibility score for every element in divisors is:
The divisibility score of divisors is 2 since nums and nums are divisible by 5.
The divisibility score of divisors is 2 since nums and nums are divisible by 7.
The divisibility score of divisors is 2 since nums and nums are divisible by 5.
Since divisors, divisors, and divisors all have the maximum divisibility score, we return the minimum of them (i.e., divisors).


Example 3:

Input: nums = , divisors = [10,16]
Output: 10
Explanation: The divisibility score for every element in divisors is:
The divisibility score of divisors is 0 since no number in nums is divisible by 10.
The divisibility score of divisors is 0 since no number in nums is divisible by 16.
Since divisors and divisors both have the maximum divisibility score, we return the minimum of them (i.e., divisors).


Constraints:

• 1 <= nums.length, divisors.length <= 1000
• 1 <= nums[i], divisors[i] <= 109

## Solution: Brute Force

Time complexity: O(m*n)
Space complexity: O(1)

## C++

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