<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) {
        long long prefixSum[nums.size() + 1];
        prefixSum[0] = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i];
        }
        deque<int>dq;
        int res = INT_MAX - 1;
        for (int i = 0; i < nums.size(); i++) {
      
            if (prefixSum[i] >= k) {
                res = min(res, i + 1);
            }
            while (!dq.empty() && prefixSum[i] - prefixSum[dq.front()] >= k) {
                res = min(res, i - dq.front());
                dq.pop_front();
            }

                      while(!dq.empty()  && prefixSum[i] <= prefixSum[dq.back()]) {
                dq.pop_back();
            }

            dq.push_back(i);
        }
           if (res == INT_MAX - 1) {
            return -1;
        } else {
            return res;
        }
    }
};
100260