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

<p>Return <em>the length of the smallest subarray that you need to remove, or </em><code>-1</code><em> if it&#39;s impossible</em>.</p>

<p>A <strong>subarray</strong> is defined as a contiguous block of elements in the array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,4,2], p = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,3,5,2], p = 9
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= p &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minSubarray(self, nums: List[int], p: int) -> int:
        target = sum(nums) % p
        if target == 0:
            return 0  # No need to remove any subarray if the sum is already divisible by p

        minLen = float("inf")
        prefix = {0: -1}  # Initialize with a base case for modulo 0
        currSum = 0

        for i, n in enumerate(nums):
            currSum = (currSum + n) % p  # Accumulate modulo p
            prefix[currSum] = i  # Store the current index for this modulo result
            required = (currSum - target) % p  # Calculate the required mod result to remove
            if required in prefix:
                minLen = min(minLen, i - prefix[required])  # Update the minimum length

        # Check if we found a valid subarray
        if minLen == float("inf"):
            return -1  # Impossible case
        if minLen == len(nums):
            return -1  # Not allowed to remove the entire array
        return minLen
        
388