<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 minInnerSubarray(self, nums: List[int], t: int) -> int:
        N = len(nums)
        curr = 0
        ans = float('inf')
        prefix = {} 
        
        for i in range(N):
            curr += nums[i]
            if curr - t in prefix:
                ans = min(ans, i - prefix[curr - t])
            prefix[curr] = i
        
        return ans if ans != float('inf') else -1
                
    def minSizeSubarray(self, nums: List[int], t: int) -> int:
        '''
        target = suffix + array * m + prefix
        t mod S = s[j] mod S + (S * m) mod S + p[i] mod S
        t = s[j] + S * m + p[i]
        To minimise the array length, we want m to be as small as possible.
        Therefore, put maximum p[i].
        Search for p[i] where p[i] mod S = (t - s[j]) mod S
        '''
        
        N = len(nums)
        S = sum(nums)
        s = [0] * N
        p = [0] * N
        modMap = {}
        curr = 0
        ans = float('inf')
        
        for i in range(N):
            curr += nums[i]
            p[i] = curr
            modMap[p[i] % S] = i
            s[i] = S - curr
        
        for j in range(N):
            mod = (t - s[j]) % S
            if mod in modMap:
                prefix = p[modMap[mod]]
                m = (t - s[j] - prefix) // S
                ans = min(ans, modMap[mod] + m * N + N - j)
    
        # Solve for single subarray
        return ans if ans != float('inf') else self.minInnerSubarray(nums, t)
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