Published at Apr 19, 2020

# Find the Smallest Divisor Given a Threshold in Python

## Problem:

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.

``````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).``````

## Constraints:

``````1 <= nums.length <= 5 * 10^4

1 <= nums[i] <= 10^6

nums.length <= threshold <= 10^6``````

## Explanation

• We need to find the smallest divisor such that the sum of nums is less than the threshold.
• The minimum value of the divisor can be 1 in that case sum of nums will remain the same whereas the maximum value of the divisor can be the maximum value in nums that will result in a minimum sum
• Basically, the value of the divisor will be in the range 1 to max(nums)
• Now we just need to check for every value in this range
• This problem can be solved by applying a binary search to get the optimal value of the divisor.

## Solution:

``````class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:
def check(mid):
result = 0
for num in nums:
result += math.ceil(num / mid)
if result > threshold:
return False
return True

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