<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();
        
//         deque<int> deq;
//         vector<long> pre(n+1);
        
//         int ans=n+1;
        
//         for(int i=0;i<n;i++)
//             pre[i+1] = pre[i] + a[i];
        
//         for(int i=0;i<n+1;i++){
            
//             while(!deq.empty() && pre[deq.back()] >= pre[i])
//                 deq.pop_back();
            
//             while(!deq.empty() && pre[i] - pre[deq.front()] >= k)
//                 ans = min(ans, i - deq.front()), deq.pop_front();
            
//             deq.push_back(i);
//         }
        
//         return ans <= n ? ans : -1;
        
        int n = a.size();
        
        vector<long> pre(n+1);
        int ans=n+1;
        
        for(int i=0;i<n;i++)
            pre[i+1] = pre[i] + a[i];
        
        auto cmp = [&](const int &i, const int &j){
                return pre[i] > pre[j];
        };
        
        priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
        
        for(int i=0;i<n+1;i++){            
            while(!pq.empty() && pre[i] - pre[pq.top()] >= k)
                ans = min(ans, i - pq.top()), pq.pop();
            
            pq.push(i);
        }
        
        return ans <= n ? ans : -1;
    }
};
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