<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:
        total_sum = sum(nums)
        n = len(nums)
        
        # If the total sum of nums is greater than or equal to target
        if total_sum >= target:
            # We can find the subarray within a single repetition
            return self.findShortestSubarray(nums, target)
        
        # Calculate the number of full repetitions needed
        full_repeats = target // total_sum
        remaining_target = target % total_sum
        
        # If remaining_target is 0, we can use full_repeats * n as the answer
        if remaining_target == 0:
            return full_repeats * n
        
        # Otherwise, find the shortest subarray for the remaining target
        min_length = self.findShortestSubarray(nums, remaining_target)
        
        # If we found a valid subarray, add the full repeats length
        if min_length != -1:
            return full_repeats * n + min_length
        
        return -1
    
    def findShortestSubarray(self, nums: List[int], target: int) -> int:
        prefix_sum = 0
        min_length = float('inf')
        prefix_map = {0: -1}  # Maps prefix_sum to its earliest index
        
        for i, num in enumerate(nums * 2):  # Consider two repetitions of nums
            prefix_sum += num
            if prefix_sum - target in prefix_map:
                min_length = min(min_length, i - prefix_map[prefix_sum - target])
            # Only store the first occurrence of each prefix_sum
            if prefix_sum not in prefix_map:
                prefix_map[prefix_sum] = i
        
        return min_length if min_length != float('inf') else -1
433