<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>

typedef long long ll;
typedef pair<ll,int> pli;
class Solution {
public:
    int shortestSubarray(vector<int>& nums, int k) {
        vector<ll> prefix(nums.size() + 1, 0);
        for(int i=1; i<prefix.size(); i++)
            prefix[i] += (prefix[i-1] + (ll)nums[i-1]);
        priority_queue<pli, vector<pli>, greater<pli>> pq;
        int minWindow = 1e9;
        for(int i=0; i<prefix.size(); i++) {
            while(pq.size() && abs(pq.top().first - prefix[i]) >= k) {
                minWindow = min(minWindow, i - pq.top().second);
                pq.pop();
            }
            pq.push({prefix[i], i});
        }
        return minWindow == 1e9 ? -1 : minWindow;
    }
};
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