# Leetcode 1283 - Find the Smallest Divisor Given a Threshold ### Understanding the problem

The problem statement on LeetCode reads as follows:

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) / 5 = 3.

Example 2:

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

Output: 3

One way to solve this problem is to use a binary search algorithm to find the smallest divisor. We can start by setting the left and right pointers to the minimum and maximum values in the array, respectively. Then, we can compute the sum of the divisions and check if it is less than or equal to the threshold. If it is, we can move the left pointer to the middle point between the left and right pointers. On the other hand, if the sum is greater than the threshold, we can move the right pointer to the middle point. We can repeat this process until the left and right pointers meet. This will give us the smallest divisor such that the sum of the divisions is less than or equal to the threshold.

Now that we have a plan, let’s implement it in Python.

``````class Solution:

def smallestDivisor(self,nums, threshold):
# Set the left and right pointers to the minimum and maximum values in the array
left, right = min(nums), max(nums)

# While the left pointer is less than the right pointer
while left < right:
# Calculate the middle point between the left and right pointers
mid = (left + right) // 2

# Calculate the sum of the divisions
sum_ = sum((n + mid - 1) // mid for n in nums)

# If the sum is less than or equal to the threshold
if sum_ <= threshold:
# Move the right pointer to the middle point
right = mid
else:
# Move the left pointer to the middle point
left = mid + 1

# Return the left pointer (which is the smallest divisor)
return left
``````