<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:
    long long pre[500000];

    int shortestSubarray(vector<int>& nums, int k) {
        int n = nums.size();
        deque<int> dq;  // Deque to store indices of the prefix sum array
        int ans = 1000000000;

        pre[0+100000] = nums[0];
        for (int i = 1; i < n; i++) {
            pre[i+100000] = nums[i] + (long long)pre[i - 1+100000];
        }

        for (int i = 0; i < n; i++) {
            if (pre[i+100000] >= k) {
                ans = min(ans, i + 1);  // If subarray from 0 to i meets the condition
            }

            // Remove indices from the front if they don't provide a shorter subarray
            while (!dq.empty() && pre[i+100000] - pre[dq.front()+100000] >= k) {
                ans = min(ans, i - dq.front());
                dq.pop_front();
            }

            // Maintain monotonicity of the deque
            while (!dq.empty() && pre[i+100000] <= pre[dq.back()+100000]) {
                dq.pop_back();
            }

            dq.push_back(i);
        }

        if (ans == 1000000000) return -1;
        return ans;
    }
};

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