<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>

typedef long long  ll;
class Solution {
public:
    int shortestSubarray(vector<int>& nums, int k) {
        ll n = nums.size();
        vector<ll>pre;
        ll sum = 0;
        for(int i = 0;i<n;i++)
        {
            sum += nums[i];
            pre.push_back(sum);
        }
        ll result = INT_MAX;

        deque<ll>dq;
        for(int i = 0;i<n;i++)
        {
            while((!dq.empty()) && (pre[dq.back()] >= pre[i]))
            {
                dq.pop_back();
            }
            if(pre[i] >= k)
            {
                result = min(result,i+1LL);
            }
            while((!dq.empty()) && ((pre[i] - pre[dq.front()]) >= k))
            {
                result = min(result,i-dq.front());
                dq.pop_front();
            }
            dq.push_back(i);
        }
        if(result == INT_MAX)
            return -1;
        return result;
    }
};
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