<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:

        for i in range(1, len(nums)):
            nums[i] = nums[i-1] + nums[i]
        
        residual = nums[-1] % p
        # print(nums, residual)

        if residual == 0:
            return 0

        # Explanation:

        # If we find a range in the array where both prefix sum % p == X
        # It would mean that the sum of elements between them added to a number % p == 0

        # Instead we want a sum of elements that would cancel our residual on top
        # Meaning that when we module the sum the reminder would be the residual
        # that we computed originally.

        # For more info look at: https://algo.monster/liteproblems/1590

        min_length = len(nums)
        seen_dict = {0: -1}
        for i in range(len(nums)):
            prefix_sum_last = nums[i]
            to_find = (prefix_sum_last - residual) % p
            if to_find in seen_dict:
                min_length = min(min_length, (i - seen_dict[to_find]))

            # If the value is present override with the most recent position since we want the smallest array
            seen_dict[prefix_sum_last % p] = i
            # print(f'i:{i} to_find:{to_find} seen_dict:{seen_dict} min_length:{min_length}')

        return min_length if min_length < len(nums) else -1
        
36600