<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>& a, int k) {
    int n = a.size();
    vector<long long> p(n + 1, 0); // prefix sum array with an extra element for handling subarray starting at index 0
    
    // Compute prefix sum
    for (int i = 0; i < n; i++) {
        p[i + 1] = p[i] + a[i];
    }

    // Deque to maintain the indices of useful prefix sums
    deque<int> dq;
    int ans = n + 1; // Set answer to a value larger than n to handle non-existent cases
    
    for (int i = 0; i <= n; i++) {
        // Ensure subarray sum >= k
        while (!dq.empty() && p[i] - p[dq.front()] >= k) {
            ans = min(ans, i - dq.front());
            dq.pop_front();
        }

        // Maintain deque for increasing prefix sums
        while (!dq.empty() && p[i] <= p[dq.back()]) {
            dq.pop_back();
        }

        dq.push_back(i);
    }

    return ans == n + 1 ? -1 : ans;
}

};
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