<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> pSum(n + 1, 0);
    
    // Compute prefix sums
    for (int i = 0; i < n; ++i) {
        pSum[i + 1] = pSum[i] + nums[i];
    }
    
    // Binary search on length of the subarray
    int left = 1, right = n + 1;
    while (left < right) {
        int mid = (left + right) / 2;
        
        bool found = false;
        deque<int> dq;
        
        // Check if there's a valid subarray of length mid
        for (int j = 0; j <= n; ++j) {
            if (!dq.empty() && j - dq.front() > mid) {
                dq.pop_front();
            }
            
            if (!dq.empty() && pSum[j] - pSum[dq.front()] >= k) {
                found = true;
                break;
            }
            
            while (!dq.empty() && pSum[dq.back()] >= pSum[j]) {
                dq.pop_back();
            }
            
            dq.push_back(j);
        }
        
        if (found) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    
    return left > n ? -1 : left;
}

};
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