# 1590. Make Sum Divisible by P

https://leetcode.com/problems/make-sum-divisible-by-p

## Description

Given an array of positive integers `nums`, remove the **smallest** subarray (possibly **empty**) such that the **sum** of the remaining elements is divisible by `p`. It is **not** allowed to remove the whole array.

Return _the length of the smallest subarray that you need to remove, or_ `-1` _if it's impossible_.

A **subarray** is defined as a contiguous block of elements in the array.

**Example 1:**

```
**Input:** nums = [3,1,4,2], p = 6
**Output:** 1
**Explanation:** The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.
```

**Example 2:**

```
**Input:** nums = [6,3,5,2], p = 9
**Output:** 2
**Explanation:** We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.
```

**Example 3:**

```
**Input:** nums = [1,2,3], p = 3
**Output:** 0
**Explanation:** Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.
```

**Example 4:**

```
**Input:** nums = [1,2,3], p = 7
**Output:** -1
**Explanation:** There is no way to remove a subarray in order to get a sum divisible by 7.
```

**Example 5:**

```
**Input:** nums = [1000000000,1000000000,1000000000], p = 3
**Output:** 0
```

**Constraints:**

* `1 <= nums.length <= 105`
* `1 <= nums[i] <= 109`
* `1 <= p <= 109`

## ac

```java
```
