<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>

#define ll long long
class Solution {
public:
    int shortestSubarray(vector<int>& nums, int k) {
        ll n = nums.size();
        set <pair<ll,ll>> s;
        s.insert({0,-1});
        ll pref = 0;
        ll ans = n+1;
        ll mx = -n-1;
        for(ll i=0; i<nums.size(); i++){
            pref += nums[i];
            s.insert({pref,i});
            while(s.begin()->first<=pref-k){
                mx = max(mx,s.begin()->second);
                s.erase(s.begin());
            }
            ans = min(ans,i-mx);
        }
        return (ans>=n+1?-1:ans);
    }
};
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