<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>>& v)
     {
        sort(v.begin(),v.end(),[](const vector<int> &a,const vector<int> &b){
            return a[0] < b[0];
        });
        bool correct = 0 ;
        while(correct == 0)
        {
         correct = 1;
         for(int i=0;i<v.size()-1;i++)
         {
            if(v[i][1] >= v[i+1][0] && v[i][0] <= v[i+1][0])
            {
                v[i][0] = min(v[i][0],v[i+1][0]);
                v[i][1] = max(v[i][1],v[i+1][1]);
                v.erase(v.begin()+i+1);
                correct = 0;
            }
         }
        }
        return v;
    }
};
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