<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:

    static bool cmp(pair<int,int> p1, pair<int,int> p2){
        if(p1.first==p2.first) return p1.second>p2.second;
        return p1.first<p2.first;
    }

    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<pair<int,int>> events;
        for(int i=0;i<intervals.size();i++){
            events.push_back(make_pair(intervals[i][0],+1));
            events.push_back(make_pair(intervals[i][1],-1));
        }
        sort(events.begin(),events.end(),cmp);
        int st,end;
        int ct=0;
        vector<vector<int>> ans;
        for(int i=0;i<events.size();i++){
            if(ct==0){
                st = events[i].first;
            }
            ct+=events[i].second;
            if(ct==0){
                end = events[i].first;
                ans.push_back({st,end});
            }
        }
        return ans;
    }
};
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