<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>& arr, int k) {
        deque<pair<int, int>> dq;
        int n = arr.size();
        int sum = 0;
        int ans = INT_MAX;
        int l = 0;
        for (int i = 0; i < n; i++) {
            sum += arr[i];
            dq.push_back(make_pair(arr[i], i));
            while (!dq.empty() && dq.back().first < 0) { // Merge
                int lastValue = dq.back().first;
                dq.pop_back();
                if (!dq.empty())
                    dq.back() = make_pair(dq.back().first + lastValue, i) ;
            }
            if (sum <= 0) {
                dq.clear();
                sum = 0;
                l = i+1;
            }
            while (!dq.empty() && sum-dq.front().first >= k) {
                int prevValue = dq.front().first;
                int index = dq.front().second;
                dq.pop_front();
                sum -= prevValue;
                l = index + 1;
            }
            if (sum >= k) {
                ans = min(ans, i - l + 1);
            }
        }
        return (ans == INT_MAX) ? -1 : ans;
    }
};
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