<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, A: List[int], p: int) -> int:
        """prefix sum remainder
        all_r = Pref[-1]
        subarray  := ［before +１, right]

        0 = Pref[-1] - Pref[right] + Pref[before]
        =>  Pref[before] == Pref[right] - Pref[-1]

        一邊hash最右邊的index
            (當前i - 先前紀錄的最右邊 = 最短)
        """
        n = len(A)
        A[0] %= p
        for i in range(1, n):
            A[i] = (A[i] + A[i - 1]) % p

        # if A[n-1] == 0: return 0

        min_len = n
        r_map_rightmost = {0: -1}
        for i in range(n):
            need_r = (A[i] - A[n - 1]) % p
            # if need_r in r_map_rightmost:
            #     min_len = min(min_len, i - r_map_rightmost[need_r])
            # r_map_rightmost[A[i]] = i
            r_map_rightmost[A[i]] = i
            if need_r in r_map_rightmost:
                min_len = min(min_len, i - r_map_rightmost[need_r])

        return min_len if min_len < n else -1

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