Question

2644. Find the Maximum Divisibility Score

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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

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

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;
  }
};

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
  }
}