<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();
        vector<long long> prefixSum(n + 1, 0); // Use long long to avoid overflow
        deque<int> q;
        int mini = INT_MAX;

        // Compute the prefix sum array
        for (int i = 1; i <= n; ++i) {
            prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
        }

        for (int i = 0; i <= n; ++i) {
            // Check if we can form a valid subarray
            while (!q.empty() && prefixSum[i] - prefixSum[q.front()] >= k) {
                mini = min(mini, i - q.front()); // Calculate the length of the subarray
                q.pop_front();
            }

            // Maintain the deque by removing indices that are not helpful
            while (!q.empty() && prefixSum[i] <= prefixSum[q.back()]) {
                q.pop_back();
            }

            // Push the current index into the deque
            q.push_back(i);
        }

        return mini == INT_MAX ? -1 : mini;
    }
};

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