<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:
        
        # Easy Approach
        n, sm, acc, cnt, ans = len(nums), sum(nums), 0, 0, []
        
        d = target//sm * n
        target%= sm
       
        nums.extend(nums)

        for i, num in enumerate(nums):

            while num > target - acc:
                acc-= nums[i - cnt]
                cnt-= 1

            acc+= num
            cnt+= 1

            if acc == target:
                ans.append(cnt + d)

        return min(ans, default = -1)

        # Another Approach ( mostly similar ) Eazy Intuitional Approach

        total_sum = sum(nums)
        n = len(nums)
        dist = float('inf') 
        t = target - (target // total_sum) * total_sum

        i = 0
        j = 0
        curr_sum = 0

        for i in range(2 * n):
            curr_sum += nums[i % n]
            while curr_sum > t:
                curr_sum -= nums[j % n]
                j += 1
            if curr_sum == t:
                dist = min(dist, i - j + 1)

        return -1 if dist == float('inf') else (target // total_sum) * n + dist
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