<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;

bool verify(vector<ll>&acc,int len,int k){
    int N=acc.size()-1;
    deque<int>que={0};
    for(int i=1;i<=N;i++){
        while(que.size()&&i-que.front()>len)que.pop_front();
        if(que.size()&&acc[i]-acc[que.front()]>=k)return true;
        while(que.size()&&acc[que.back()]>acc[i])que.pop_back();
        que.push_back(i);
    }
    return false;
}


class Solution {
public:
    int shortestSubarray(vector<int>& nums, int k) {
        int N=nums.size();
        nums.insert(nums.begin(),0);
        vector<ll>acc(N+1,0);
        for(int i=1;i<=N;i++)acc[i]=acc[i-1]+nums[i];
        int lft=0,rht=N;
        while(lft<rht){
            int mid=lft+((rht-lft)>>1);
            if(verify(acc,mid,k))rht=mid;
            else lft=mid+1;
        }
        return verify(acc,lft,k)?lft:-1;
    }
};
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