<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>

using LL = long long;
class Solution {
public:
    int shortestSubarray(vector<int>& nums, int k) {
        nums.insert(nums.begin(), 0);
        int n = nums.size();
        LL prefix_sum[n];
        prefix_sum[0] = nums[0];
        for(int i=1; i<n; i++) prefix_sum[i] = prefix_sum[i-1] + nums[i];
        /*
              find   
            [xxxxxj] [xxxxi]
            [xxxxxj   xxxxi] prefix_sum
            prefix_sum - find >= k
        */
        int min_len = INT_MAX;
        deque<int> dq;
        for(int i=0; i<n; i++){
            while(!dq.empty() && prefix_sum[dq.back()]>=prefix_sum[i]) dq.pop_back();
            while(!dq.empty() && prefix_sum[i]-prefix_sum[dq.front()]>=k){
                min_len = min(min_len, i-dq.front());
                dq.pop_front();
            }
            dq.push_back(i);
        }
        return min_len==INT_MAX ?-1 :min_len;
    }
};
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