leetcode1283. Find the Smallest Divisor Given a Threshold

Posted seyjs

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了leetcode1283. Find the Smallest Divisor Given a Threshold相关的知识,希望对你有一定的参考价值。

题目如下:

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

解题思路:除数越大,计算出来的和越小,因此可以使用二分查找法。

代码如下:

class Solution(object):
    def smallestDivisor(self, nums, threshold):
        """
        :type nums: List[int]
        :type threshold: int
        :rtype: int
        """
        res = 0
        low,high = 1,1000000
        while low <= high:
            mid = (low + high)/2
            count = 0
            for i in nums:
                count += (i/mid)
                if i%mid != 0:count += 1
            if count <= threshold:
                res = mid
                high = mid - 1
            else:
                low = mid + 1
        return res

以上是关于leetcode1283. Find the Smallest Divisor Given a Threshold的主要内容,如果未能解决你的问题,请参考以下文章

LeetCode-第 166 场周赛

LeetCode 564. Find the Closest Palindrome

#Leetcode# 997. Find the Town Judge

LeetCode: Find the Duplicate Number

LeetCode 997. Find the Town Judge

Leetcode: Find the Difference