<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:
        left=0
        right=0
        count=0
        heap=[]
        span=inf
    
        for i in range(len(nums)):
            count+=nums[i]
            temp=[]
            if count<=0:
                count=0
                left=i+1
                right=i+1
                heap=[]
            else:
                right=i+1
                heappush(heap,(count,right))
            temp_count=count
            while temp_count>=k:
                if len(temp):
                    heappop(heap)
                    left=temp[1]
                span=min(span,right-left)
                temp_count=count
                temp=heap[0]
                if left<temp[1]:
                    temp_count-=temp[0]
            
        if span==inf:
            return -1
        else:
             return span
                     
                
34000