<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:
    #define ll long long 
    
    int shortestSubarray(vector<int>& nums, int k) {
           vector<int>v = nums;
           ll n = v.size();
           ll sum = 0;
           ll minu = 1e9;
           priority_queue<pair<ll,ll>,vector<pair<ll,ll> >,greater<pair<ll,ll> > >pq;
           for(ll i = 0;i<n;i++){
            sum +=v[i];
            if(sum>=k){
                minu = min(minu,i+1);
            }
            while(!pq.empty()&&sum-pq.top().first>=k){
                minu = min((ll)minu,i-pq.top().second);
                pq.pop();
            }
            pq.push({sum,i}); 
           }
           if(minu==1e9){
            return -1;
           }
           return minu;
           
    }
};
123260