<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 {
public:
    int shortestSubarray(vector<int>& nums, int k) 
    {
        int len = INT_MAX;
        long long sum = 0;
        vector<int> positive;
        for (int i = 0, j = 0; j < nums.size(); ++j) 
        {
            sum += nums[j];
            if (nums[j] < 0) 
            {
                int r = nums[j];
                while (!positive.empty()) 
                {
                    int cur = positive.back();
                    r += nums[cur];
                    nums[cur] = r;
                    nums[j] = 0;
                    positive.pop_back();

                    if (r < 0 && !positive.empty() && positive.back() >= i) 
                        nums[cur] = 0;
                    
                    else 
                    {
                        positive.push_back(cur);
                        break;
                    }
                }
            }
            else 
            {
                positive.push_back(j);

                while (i <= j && (sum-nums[i] >= k || nums[i] <= 0))
                    sum -= nums[i++];
                
                if (sum >= k) 
                    len = min(len, j - i + 1);
            }
        }
        return len == INT_MAX ? -1 : len;
    }
};
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