<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:
        # some variation on hashmap prefix trick
        # look for some min subarray of nums (standard)
        # also look for max of sum(nums) - target, the longest subarray that isn't target
        total = sum(nums)
        
        # if target is much larger than the total array, we go around several times
        multiple = 0
        if target == total:
            return len(nums)
        elif target > total:
            multiple = target // total  # multiply this with answer later
            target = target % total # the remaining nums to look for
            
            if target == 0:
                # multiple of nums, just return
                return multiple * len(nums)
        
        inv_target = total - target
        
        prefix = 0
        
        mp = defaultdict(int)
        mp[0] = -1
        best = float('inf')

        mp_inv = defaultdict(int)
        mp[0] = -1
        best_inv = 0
        
        for r, n in enumerate(nums):
            prefix += n

            if (prefix - target) in mp:
                l = mp[prefix - target]
                best = min(best, r - l)
                
            if (prefix - inv_target) in mp_inv:
                l = mp_inv[prefix - inv_target]
                best_inv = max(best_inv, r - l)

            mp[prefix] = r
            
            if prefix not in mp_inv:
                mp_inv[prefix] = r
         
        if best == float('inf') and best_inv == 0:
            return -1 # no match either forward or back
        
        return multiple * len(nums) + min(best, len(nums) - best_inv)
                
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