[leetcode] 1283. Find the Smallest Divisor Given a Threshold

Description

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Example 3:

Input: nums = [19], threshold = 5
Output: 4

Constraints:

• 1 <= nums.length <= 5 * 10^4
• 1 <= nums[i] <= 10^6
• nums.length <= threshold <= 10^6

分析

题目的意思是：这道题需要找出一个最小的数x，x除以数组nums中所有的数的商要小于等于给定的阈值，题目中的商取上界。思路直接，用二分法找出满足条件的数就行了，二分法的上界当然就是数组的和了。我看了一下答案，跟我的思路差不多，只是找上界有点不一样，不过还好。

代码

class Solution:
def solve(self,nums,val):
res=0
for num in nums:
res+=(num+val-1)//val
return res
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
l=1
r=sum(nums)
while(l<r):
mid=l+(r-l)//2
if(self.solve(nums,mid)>threshold):
l=mid+1
else:
r=mid
return l

参考文献

[LeetCode] Solution

最后

以上就是暴躁寒风为你收集整理的[leetcode] 1283. Find the Smallest Divisor Given a Threshold的全部内容，希望文章能够帮你解决[leetcode] 1283. Find the Smallest Divisor Given a Threshold所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。