<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 n=nums.size();
        long long lo=1,hi=n+1,mid,i,s[n+1];
        s[0]=0;
        for(i=0;i<n;i++) s[i+1]=s[i]+nums[i];
        while(hi-lo>1){
            mid=(hi+lo)/2;
            cout<<mid<<endl;
            deque<int> dq;
            dq.push_back(0);
            for(i=0;i<nums.size();i++){
                while((!dq.empty()) && dq.front()<=i-mid) dq.pop_front();
                if((!dq.empty()) && s[i+1]-s[dq.front()]>=k) break;
                // if((dq.empty()) && s[i+1]>=k) break;
                while((!dq.empty()) && s[i+1]<=s[dq.back()]) dq.pop_back();
                dq.push_back(i+1);
            }
            if(i==nums.size()) lo=mid+1;
            else hi=mid;
        }
            cout<<lo<<" "<<hi<<endl;
            deque<int> dq;
            dq.push_back(0);
            for(i=0;i<nums.size();i++){
                while((!dq.empty()) && dq.front()<=i-lo) dq.pop_front();
                if((!dq.empty()) &&s[i+1]-s[dq.front()]>=k) break;
                // if((dq.empty()) && s[i+1]>=k) break;
                while((!dq.empty()) &&s[i+1]<=s[dq.back()]) dq.pop_back();
                dq.push_back(i+1);
            }
            
            if(i<nums.size()) return lo;

            dq.clear();
            dq.push_back(0);
            for(i=0;i<nums.size();i++){
                while((!dq.empty()) && dq.front()<=i-hi) dq.pop_front();
                if((!dq.empty()) &&s[i+1]-s[dq.front()]>=k) break;
                // if((dq.empty()) && s[i+1]>=k) break;
                while((!dq.empty()) &&s[i+1]<=s[dq.back()]) dq.pop_back();
                dq.push_back(i+1);
            }
            if(i<nums.size()) return hi;
            return -1;
            
    }
};
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