human
I found an interesting problem on 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). 

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

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python program to find the smallest divisor such that the result mentioned above is less than or equal to threshold.

class Solution:
    def smallestDivisor(self, nums: List[int], threshold: int) -> int:
        def check(div):
            # print(sum([elem/div if elem % div == 0 else (elem//div)+1 for elem in nums]))
            return threshold >= sum([elem/div if elem % div == 0 else (elem//div)+1 for elem in nums])
        
        def bst(low,high):
            # print(low,high)
            mid = (low+high) // 2
            if low > high:
                return -1
            elif low == high:
                if check(low):
                    return low
                else:
                    return -1
            else:
                if check(mid):
                    ambition = bst(low,mid-1)
                    if ambition != -1:
                        return ambition
                    else:
                        return mid
                else:
                    return bst(mid+1,high)
                
        return bst(1,10**9)