<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>

#include <bits/stdc++.h>

class Solution {
public:
    int min(int a,int b) {
        if(a>b)
            return b; 

        return a;
    }
    int shortestSubarray(vector<int>& nums, int k) {
        
        int n = nums.size() ; 
        int ls = n+1; 
        deque<long long> a; 
        long long x[n];
        x[0]=nums[0]; 
        for(int i=1;i<n ;i++) {
            x[i] = x[i-1] + nums[i] ;
        } 
        for(int i=0; i<n ;i++ ){ 
            
            if(x[i]>=k) {
                ls=min(i+1,ls) ;
            }
            while(a.size()>0 && x[i]-x[a.front()]>=k) {
                ls = min(ls,i-a.front());  
                a.pop_front() ;
            } 
            while(a.size()>0 && x[i]<=x[a.back()]) { 
                a.pop_back(); 
            }
            a.push_back(i); 
        }

        return ls<=n? ls : -1 ; 

        
    }
};
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