<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, target):
        n = len(nums)
        total_sum = sum(nums)
        
        # If the target is greater than or equal to the total sum,
        # we can reduce the problem to a single period of nums.
        if target >= total_sum:
            period_count = target // total_sum
            target %= total_sum
            if target == 0:
                return period_count * n
        else:
            period_count = 0
        
        # Sliding window approach
        window_sum = 0
        min_length = float('inf')
        left = 0
        
        for right in range(n * 2):
            window_sum += nums[right % n]
            
            # Shrink the window from the left if the sum exceeds the target
            while window_sum > target:
                window_sum -= nums[left % n]
                left += 1
            
            # Update the minimum length if the current sum equals the target
            if window_sum == target:
                min_length = min(min_length, right - left + 1)
        
        return period_count * n + min_length if min_length != float('inf') else -1

# Example usage:
solution = Solution()
nums = [1, 2, 3]
target = 5
print(solution.minSizeSubarray(nums, target))  # Output: 2
384