<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, p):
        # Calculate total sum of the array
        total_sum = sum(nums)
        mod_needed = total_sum % p  # remainder when divided by p
        
        # If already divisible by p, return 0 (no need to remove anything)
        if mod_needed == 0:
            return 0
        
        prefix_sum = 0  # to track sum as we move through the array
        mod_map = {0: -1}  # Hash map to store (prefix_sum % p) and index, with base case
        min_len = len(nums)  # initialize the minimum length with array size
        
        for i, num in enumerate(nums):
            # Accumulate prefix sum
            prefix_sum += num
            # Current mod of the prefix sum
            current_mod = prefix_sum % p
            
            # We want to find if there's a prefix we can remove such that the remaining sum is divisible by p
            target_mod = (current_mod - mod_needed) % p
            
            # If we have seen this target mod before, calculate the subarray length
            if target_mod in mod_map:
                min_len = min(min_len, i - mod_map[target_mod])
            
            # Update the map with the current mod and index
            mod_map[current_mod % p] = i
        
        # If we couldn't find any subarray to remove, return -1
        return min_len if min_len < len(nums) else -1

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