题目 #
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
- $1 <= nums[i], divisors[i] <= 10^9$
思路1 #
分析 #
- 照着做,爆破
代码 #
1func maxDivScore(nums []int, divisors []int) int {
2 res, rc := math.MaxInt, -1
3 for _, v := range divisors {
4 cnt := 0
5 for _, v1 := range nums {
6 if v1%v == 0 {
7 cnt++
8 }
9 }
10 if cnt > rc || (rc == cnt && res > v) {
11 res, rc = v, cnt
12 }
13 }
14 return res
15}