<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) {
        int l = 0, r = 0, sum = 0, ans = INT_MAX, neg = 0;
        while(r < nums.size()){
            if(sum + nums[r] > 0){
                if(nums[r] < 0)neg -= nums[r];
                sum += nums[r];
            }
            else{
                neg = 0;
                sum = 0;
                l = r + 1;
            }
            if(sum >= k){
                int tmp = sum, tmp_idx = l;
                while(tmp_idx <= r){
                    if(tmp >= k){
                        ans = min(ans, r - tmp_idx + 1);
                        sum = tmp;
                        l = tmp_idx;
                    }else{
                        if(tmp + neg < k)break;
                    }
                    tmp -= nums[tmp_idx++];
                }
            }
            r++;
        }
        return (ans == INT_MAX)? -1 : ans;
    }
};
95660