<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 N = nums.size(), res = N + 1;
        deque<long> d;
        long long A[N];
        for(int i=0;i<N;i++){
            A[i]=nums[i];
        }
        for (int i = 0; i < N; i++) {
            if (i > 0)
                A[i] += A[i - 1];
            if (A[i] >= K)
                res = min(res, i + 1);
            while (d.size() > 0 && A[i] - A[d.front()] >= K)
                {int c=i-d.front();
                    res = min(res, c);
                d.pop_front();}
            while (d.size() > 0 && A[i] <= A[d.back()])
                d.pop_back();
            d.push_back(i);
        }
        return res <= N ? res : -1;
    } 
};
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