<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>& nums, int k) {
        if (nums.empty())
            return -1;

        int n = nums.size();
        int len = INT_MAX;
        deque<pair<long,int>>dq;
        vector<int>prefix(n+1);
        long sum = 0;
        prefix[0] = 1;
        for (int i = 1; i<=n; i++)
            prefix[i] += nums[i-1];
        
        for (int i=0; i<n; i++) {
            sum += nums[i];
            if (sum >= k)
                len = min(len, i+1);
            while (!dq.empty() && sum <= dq.back().first)
                dq.pop_back();
            while (!dq.empty() && sum - dq.front().first >= k) {
                len = min(len, i-dq.front().second);
                dq.pop_front();
            }
            dq.push_back({sum, i});
        }
        return len == INT_MAX ? -1 : len;
        
    }
};
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