Welcome to Subscribe On Youtube
2644. Find the Maximum Divisibility Score
Description
You are given two 0indexed 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] <= 10^{9}
Solutions

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; } }

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; } };

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

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 }

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; }

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 } }