<p>You are given a <strong>0-indexed</strong> array <code>nums</code> and an integer <code>target</code>.</p>

<p>A <strong>0-indexed</strong> array <code>infinite_nums</code> is generated by infinitely appending the elements of <code>nums</code> to itself.</p>

<p>Return <em>the length of the <strong>shortest</strong> subarray of the array </em><code>infinite_nums</code><em> with a sum equal to </em><code>target</code><em>.</em> If there is no such subarray return <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3], target = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example infinite_nums = [1,2,3,1,2,3,1,2,...].
The subarray in the range [1,2], has the sum equal to target = 5 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,2,3], target = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example infinite_nums = [1,1,1,2,3,1,1,1,2,3,1,1,...].
The subarray in the range [4,5], has the sum equal to target = 4 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,6,8], target = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> In this example infinite_nums = [2,4,6,8,2,4,6,8,...].
It can be proven that there is no subarray with sum equal to target = 3.
</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>5</sup></code></li>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minSizeSubarray(self, nums: List[int], target: int) -> int:
        n = len(nums)

        pfx_sum = {0: -1}
        shortest = inf
        
        # current = 0
        # for i in range(2 * n):
        #     current += nums[i % n]
        #     if current - target in pfx_sum:
        #         shortest = min(shortest, i - pfx_sum[current - target])
        #     if i < n:
        #         pfx_sum[current] = i

        # for k in range(1, n + 1):
        #     for i in range(n):
        #         current += nums[i]
        #         if current - target in pfx_sum:
        #             shortest = min(shortest, i + k * n - pfx_sum[current - target])
        #         pfx_sum[current] = i + k * n

        # return shortest if shortest != float('inf') else -1

        window, target = divmod(target, sum(nums))
        if target == 0: 
            return n * window
        
        current = 0
        for i, num in enumerate(2 * nums):
            current += num
            if current - target in pfx_sum:
                shortest = min(shortest, i - pfx_sum[current - target])
            pfx_sum[current] = i

        return (n * window) + shortest if shortest != inf else -1
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