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