2644. Find the Maximum Divisibility Score
Description
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 <= 10001 <= 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 $O(1)$.
Python3
class Solution:
def maxDivScore(self, nums: List[int], divisors: List[int]) -> int:
ans, mx = divisors[0], 0
for div in divisors:
cnt = sum(x % div == 0 for x in nums)
if mx < cnt:
mx, ans = cnt, div
elif mx == cnt and ans > div:
ans = div
return ans
Java
class Solution {
public int maxDivScore(int[] nums, int[] divisors) {
int ans = divisors[0];
int mx = 0;
for (int div : divisors) {
int cnt = 0;
for (int x : nums) {
if (x % div == 0) {
++cnt;
}
}
if (mx < cnt) {
mx = cnt;
ans = div;
} else if (mx == cnt) {
ans = Math.min(ans, div);
}
}
return ans;
}
}
C++
class Solution {
public:
int maxDivScore(vector<int>& nums, vector<int>& divisors) {
int ans = divisors[0];
int mx = 0;
for (int div : divisors) {
int cnt = 0;
for (int x : nums) {
cnt += x % div == 0;
}
if (mx < cnt) {
mx = cnt;
ans = div;
} else if (mx == cnt) {
ans = min(ans, div);
}
}
return ans;
}
};
Go
func maxDivScore(nums []int, divisors []int) int {
ans, mx := divisors[0], 0
for _, div := range divisors {
cnt := 0
for _, x := range nums {
if x%div == 0 {
cnt++
}
}
if mx < cnt {
ans, mx = div, cnt
} else if mx == cnt && ans > div {
ans = div
}
}
return ans
}
TypeScript
function maxDivScore(nums: number[], divisors: number[]): number {
let ans: number = divisors[0];
let mx: number = 0;
for (const div of divisors) {
const cnt = nums.reduce((a, b) => a + (b % div == 0 ? 1 : 0), 0);
if (mx < cnt) {
mx = cnt;
ans = div;
} else if (mx === cnt && ans > div) {
ans = div;
}
}
return ans;
}
Rust
impl Solution {
pub fn max_div_score(nums: Vec<i32>, divisors: Vec<i32>) -> i32 {
let mut ans = divisors[0];
let mut mx = 0;
for &div in &divisors {
let mut cnt = 0;
for &n in &nums {
if n % div == 0 {
cnt += 1;
}
}
if cnt > mx || (cnt >= mx && div < ans) {
mx = cnt;
ans = div;
}
}
ans
}
}