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