<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<pair<int,int>> temp;

        for(auto v:intervals){
            pair<int,int> p = {v[0],v[1]};
            temp.push_back(p);
        }
        sort(temp.begin(),temp.end());
        vector<vector<int>> ans;
        
        stack<pair<int,int>> st;

        st.push(temp[0]);

        for(int i=1;i<temp.size();i++){
            pair<int,int> p = st.top();
            if(p.second>=temp[i].first){
                p = {p.first,max(p.second,temp[i].second)};
                st.pop();
                st.push(p);
            }
            else{
                st.push(temp[i]);
            }
        }
        vector<vector<int>> answer;
        while(!st.empty()){
            pair<int,int> p = st.top();
            st.pop();
            answer.push_back({p.first,p.second});
        }
        reverse(answer.begin(),answer.end());
        return answer;
    }
};
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