<p>Given an array of positive integers <code>nums</code> and a positive integer <code>target</code>, return <em>the <strong>minimal length</strong> of a </em><span data-keyword="subarray-nonempty"><em>subarray</em></span><em> whose sum is greater than or equal to</em> <code>target</code>. If there is no such subarray, return <code>0</code> instead.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> target = 7, nums = [2,3,1,2,4,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The subarray [4,3] has the minimal length under the problem constraint.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> target = 4, nums = [1,4,4]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> target = 11, nums = [1,1,1,1,1,1,1,1]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> If you have figured out the <code>O(n)</code> solution, try coding another solution of which the time complexity is <code>O(n log(n))</code>.
class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        min_length = n + 1
        left = 0
        curr_sum = 0

        for right in range(n):
            curr_sum += nums[right]
            while curr_sum - nums[left] >= target:
                curr_sum -= nums[left]
                left += 1
            if curr_sum >= target:
                min_length = min(min_length, right - left + 1)

            print(curr_sum, min_length, left, right)

        if min_length == n + 1:
            return 0
        return min_length
        

            
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