<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) {
        
      if(intervals.size()==1)
         return intervals;
      vector<pair<int,int>> p;
      for(int i=0;i<intervals.size();i++)
      {
          p.push_back({intervals[i][0],intervals[i][1]});
      } 
      sort(p.begin(),p.end());

      vector<vector<int>> ans;
      int f=p[0].first,s=p[0].second;
      for(int i=0;i<p.size()-1;i++)
      {
          vector<int> a(2);
          if(s>=p[i+1].first)
          {
              s=max(s,p[i+1].second);
          }
          else
          {
              a[0]=f;
              a[1]=s;
              f=p[i+1].first;
              s=p[i+1].second;
              ans.push_back(a);
          }
      } 
      int n=intervals.size();
      ans.push_back({f,s});
      return ans;
    }
};
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