<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)
        
        # If the total sum is greater than or equal to the target, we can use the wrap-around property
        # Calculate how many full cycles of nums we need to reach or exceed the target
        full_cycles = target // total_sum
        remainder_target = target % total_sum
        
        # If remainder_target is 0, it means we can achieve the target with full cycles of nums
        if remainder_target == 0:
            return full_cycles * n
        
        # Use a dictionary to store prefix sums and their indices
        prefix_sum = 0
        prefix_sum_map = {0: -1}
        min_length = float('inf')
        
        # Iterate over the array twice to simulate the infinite array
        for i in range(2 * n):
            prefix_sum += nums[i % n]
            
            # Check if there is a subarray that sums to remainder_target
            if (prefix_sum - remainder_target) in prefix_sum_map:
                length = i - prefix_sum_map[prefix_sum - remainder_target]
                min_length = min(min_length, length)
            
            # Store the prefix sum and its index
            if prefix_sum not in prefix_sum_map:
                prefix_sum_map[prefix_sum] = i
        
        # If we found a valid subarray, return its length plus the full cycles
        if min_length != float('inf'):
            return min_length + full_cycles * n
        
        # If no valid subarray is found, return -1
        return -1
57700