# Find the Smallest Divisor Given a Threshold

{% hint style="info" %}
<https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/>
{% endhint %}

{% tabs %}
{% tab title="Question" %}
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). 
```

{% endtab %}

{% tab title="Answer" %}

```python
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.
{% endtab %}
{% endtabs %}


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