<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the length of the shortest non-empty <strong>subarray</strong> of </em><code>nums</code><em> with a sum of at least </em><code>k</code>. If there is no such <strong>subarray</strong>, return <code>-1</code>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> 1
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,2], k = 4
<strong>Output:</strong> -1
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [2,-1,2], k = 3
<strong>Output:</strong> 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>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def shortestSubarray(self, nums: List[int], k: int) -> int:
        n = len(nums)
        minLen = n + 1

        stack = [0]
        prefixSum = [0] * (n + 1)
        left = 0
        for i in range(n):
            prefixSum[i+1] = prefixSum[i] + nums[i]
            while stack and prefixSum[i+1] <= prefixSum[stack[-1]]:
                if left == len(stack) - 1 and left != 0:
                    left -= 1
                stack.pop(-1)
            stack.append(i+1)
            #print(f'i = {i} prefixSum = {prefixSum} stack = {stack} left = {left}')

            while left < len(stack) and prefixSum[i+1] >= prefixSum[stack[left]] + k:
                #print(f'prefixSum[i+1] = {prefixSum[i+1]} prefixSum[stack[left]] = {prefixSum[stack[left]]} left = {left}')
                
                minLen = min(minLen, i + 1 - stack[left])
                left += 1

        return -1 if minLen == n + 1 else minLen
32500