<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>

from typing import List

class Solution:
    def minSizeSubarray(self, nums: List[int], target: int) -> int:
        n = len(nums)
        total_sum = sum(nums)
        
        # Calculate how many full cycles of nums are needed to reach the target
        full_cycles_needed = target // total_sum
        
        # Remaining sum needed after full cycles
        remaining = target - full_cycles_needed * total_sum
        
        # If the remaining sum is zero, then we can exactly use full_cycles_needed * n to form the target
        if remaining == 0:
            return full_cycles_needed * n

        # We're going to simulate checking subarrays of the doubled array
        # by considering indices up to 2n (since the input length is 1 <= n <= 10^5)
        extended_nums = nums + nums
        min_len = float('inf')
        
        # Use a sliding window to find the shortest subarray of `extended_nums`
        current_sum = 0
        start = 0
        
        for end in range(2 * n):
            current_sum += extended_nums[end]
            
            while current_sum > remaining and start <= end:
                current_sum -= extended_nums[start]
                start += 1
            
            if current_sum == remaining:
                # The end index might be in the second cycle, adjust subarray length by current cycle
                subarray_length = end - start + 1
                min_len = min(min_len, subarray_length)

        # If we have found such a valid subarray
        if min_len != float('inf'):
            return full_cycles_needed * n + min_len
        
        return -1
370