<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) {
        int n = nums.size();
        deque<pair<long long, long long>> dq; //index, sum
        long long sum = 0;
        long long shortest = INT_MAX;

        for(long long i = 0; i < n; ++i) {
            sum += nums[i];
            if(sum >= k) shortest = min(shortest, i + 1); // Sum from start to the i-th index

            // Reduce window size to find minimum window with sum>=k
            pair<long long, long long> curr = {INT_MIN, INT_MIN};
            while(!dq.empty() && (sum - dq.front().second) >= k) {
                curr = dq.front();
                dq.pop_front();
            }

            // Calculate new shortest (if possible)
            if(curr.second != INT_MIN) {
                shortest = min(shortest, (i - curr.first));
            }

            // Maintain monotonically non-decreasing order of deque
            while(!dq.empty() && sum <= dq.back().second) {
                dq.pop_back();
            }
            dq.push_back({i, sum}); // Push the i-th sum
        }
        return shortest == INT_MAX ? -1 : shortest;
    }
};
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