<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) {
        long long res = LLONG_MAX;

        vector<long long> p(nums.size() + 1, 0);
        deque<long long> dq;
        dq.push_back(0);

        int r = 0;

        for (int i = 0; i < nums.size(); i++) {
            p[i + 1] = p[i] + nums[i];

            while (!dq.empty() && p[dq.back()] > p[i + 1]) {
                dq.pop_back();
            }
            dq.push_back(i + 1);

            while (p[i + 1] - p[dq.front()] >= k) {
                res = min(res, static_cast<long long>(i - r + 1));
                if (dq.front() == r) {
                    dq.pop_front();
                }
                r++;
            }
        }

        return res == LLONG_MAX ? -1 : static_cast<int>(res);
    }
};
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