<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) {
        vector<pair<long long, size_t> > stk;
        long long curSum = 0;
        int ans = -1;
        stk.push_back({0, 0});
        for (size_t i = 0; i < nums.size(); ++i) {
            curSum += nums[i];
            auto it = std::upper_bound(stk.begin(), stk.end(), make_pair(curSum - k, nums.size()));
            if (it != stk.begin()) {
                --it;
                if (ans == -1 || i + 1 - it->second < ans) {
                    ans = i + 1 - it->second;
                }
            }
            while (stk.size() && stk.back().first >= curSum) {
                stk.pop_back();
            }
            stk.push_back({curSum, i+1});
        }
        return ans;
    }
};
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