<p>Given an array&nbsp;of <code>intervals</code>&nbsp;where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, merge all overlapping intervals, and return <em>an array of the non-overlapping intervals that cover all the intervals in the input</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[1,3],[2,6],[8,10],[15,18]]
<strong>Output:</strong> [[1,6],[8,10],[15,18]]
<strong>Explanation:</strong> Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[1,4],[4,5]]
<strong>Output:</strong> [[1,5]]
<strong>Explanation:</strong> Intervals [1,4] and [4,5] are considered overlapping.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        
        using pii = pair<int, int>;
        priority_queue<pii, vector<pii>, greater<pii>> pq;
        for(int i = 0; i < intervals.size(); i++) {
            const auto& interval = intervals[i];
            pq.push({interval[0], interval[1]});
        }

        vector<vector<int>> res;
        while(!pq.empty()) {

            if(pq.size() > 1) {

                auto p1 = pq.top();
                pq.pop();
                auto p2 = pq.top();
                pq.pop();

                // p1 = [1, 3], p2 = [2, 6]

                if(p2.first <= p1.second) { // overlap
                    pq.push({p1.first, max(p2.second, p1.second)});
                }else{ // no overlap
                    res.push_back({p1.first, p1.second});
                    pq.push({p2.first, p2.second});
                }

            }else{
                auto p = pq.top();
                pq.pop();
                res.push_back({p.first, p.second});
            }


        }
        
        return res;
    }
};
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