Find the Smallest Divisor Given a Threshold

https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/

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).

class Solution:
    def smallestDivisor(self, nums: List[int], threshold: int) -> int:
        def condition(divisor) -> bool:
            return sum((num - 1) // divisor + 1 for num in nums) <= threshold

        left, right = 1, max(nums)
        while left < right:
            mid = left + (right - left) // 2
            if condition(mid):
                right = mid
            else:
                left = mid + 1
        return left

After so many problems introduced above, this one should be a piece of cake. We don't even need to bother to design a condition function, because the problem has already told us explicitly what condition we need to satisfy.

PreviousUgly Number 3 NextKth smallest number in multiplication table

Last updated 5 years ago