<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>& num, int k) {
        int n = num.size();
        list<long> l;
        vector<long> nums(n);
        for(int i = 0 ; i<n ; i++) nums[i] = num[i];

        int ans = INT_MAX;

        for(int i = 0 ; i<n ; i++)
        {
            if(i > 0) nums[i] += nums[i - 1];

            if(nums[i] >= k) ans = min(ans , i + 1);

            while(l.size() && nums[i] - nums[l.front()] >= k)
            {
                ans = min( ans , i - (int)l.front());
                l.pop_front();
            }

            // just increasing in the number 
            while(l.size() && nums[i] < nums[l.back()]) l.pop_back();

            l.push_back(i);
        }

        return ans == INT_MAX ? -1 : ans;
    }
};
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