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Java-based LeetCode algorithm problem solutions, regularly updated
1590\. Make Sum Divisible by P
Medium
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.
**Constraints:**
* 1 <= nums.length <= 105
* 1 <= nums[i] <= 109
* 1 <= p <= 109