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# 1283. Find the Smallest Divisor Given a Threshold

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

## Description

Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer `divisor`, divide all the array by it, and sum the division's result. Find the **smallest** `divisor` such that the result mentioned above is less than or equal to `threshold`.

Each result of the 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 of 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2). 
```

**Example 2:**

```
**Input:** nums = [44,22,33,11,1], threshold = 5
**Output:** 44
```

**Example 3:**

```
**Input:** nums = [21212,10101,12121], threshold = 1000000
**Output:** 1
```

**Example 4:**

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

**Constraints:**

* `1 <= nums.length <= 5 * 104`
* `1 <= nums[i] <= 106`
* `nums.length <= threshold <= 106`

## ac

```java
```


---

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