<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>& A, int k) {
        int n=A.size();
        vector<long>nums(n,A[0]);
        deque<long>dq;
        int ans=INT_MAX;
        for(int i=0;i<n;i++){
            if(i>0) nums[i]=nums[i-1]+A[i];
            if(nums[i]>=k) ans=min(ans,i+1);
            while(!dq.empty() && nums[i]-nums[dq.front()] >=k){
                ans=min(ans,i-(int)dq.front());
                dq.pop_front();
            }
            while(!dq.empty() && nums[i]<=nums[dq.back()]){
                dq.pop_back();
            }
            dq.push_back(i);
        }
        return ans==INT_MAX?-1:ans;
    }
};
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