2644. Find the Maximum Divisibility Score
Description
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[0] is 0 since no number in nums is divisible by 5. The divisibility score of divisors[1] is 1 since 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. Since divisors[2] 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[0] is 2 since nums[0] and nums[3] are divisible by 5. The divisibility score of divisors[1] is 2 since nums[1] and nums[2] are divisible by 7. The divisibility score of divisors[2] is 2 since nums[0] and nums[3] are divisible by 5. Since divisors[0], divisors[1], and divisors[2] all have the maximum divisibility score, we return the minimum of them (i.e., divisors[2]).
Example 3:
Input: nums = [12], divisors = [10,16] Output: 10 Explanation: The divisibility score for every element in divisors is: The divisibility score of divisors[0] is 0 since no number in nums is divisible by 10. The divisibility score of divisors[1] is 0 since no number in nums is divisible by 16. Since divisors[0] and divisors[1] both have the maximum divisibility score, we return the minimum of them (i.e., divisors[0]).
Constraints:
1 <= nums.length, divisors.length <= 1000
1 <= nums[i], divisors[i] <= 109
Solutions
Solution 1: Enumeration
We can enumerate each element $div$ in $divisors$, and calculate how many elements in $nums$ can be divided by $div$, denoted as $cnt$.
- If $cnt$ is greater than the current maximum divisibility score $mx$, then update $mx = cnt$, and update $ans = div$.
- If $cnt$ equals $mx$ and $div$ is less than $ans$, then update $ans = div$.
Finally, return $ans$.
The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of $nums$ and $divisors$ respectively. The space complexity is $O(1)$.
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