<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) {
        vector<vector<int>> ans;
        int n = intervals.size();
        vector<int> ni(2);
        int i=0;

        sort(intervals.begin(), intervals.end());
        for(auto it: intervals)
        for(auto j : it)
        cout<< j;
        cout<< " ";

        if(n == 1){
            ans.push_back(intervals[i]);
            return ans;
        }

        while(i<n){
            if(i == n-1){
                ans.push_back(intervals[i]);
                return ans;
            }
            ni[0] = intervals[i+1][0];
            ni[1] = intervals[i+1][1];
            
            while(i<n && intervals[i][1] < ni[0]){
                ans.push_back(intervals[i]);
                i++;
            }
            while(i<n && intervals[i][0] <= ni[1]){
                ni[0] = min(ni[0],intervals[i][0]);
                ni[1] = max(ni[1], intervals[i][1]);
                i++;
            }
            ans.push_back(ni);
        }
        return ans;
    }
};
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