<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:
        l = target // sum(nums)
        target %= sum(nums)
        if target == 0:
            return l * len(nums)

        prev = {0: -1}
        acc = 0
        res = inf
        for i in range(len(nums) * 2):
            acc += nums[i % len(nums)]
            if acc - target in prev:
                res = min(res, i - prev[acc - target])
            prev[acc] = i
        return res + l * len(nums) if res != inf else -1

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