<p>Given an array of intervals <code>intervals</code> where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, return <em>the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[1,2],[2,3],[3,4],[1,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> [1,3] can be removed and the rest of the intervals are non-overlapping.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[1,2],[1,2],[1,2]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> You need to remove two [1,2] to make the rest of the intervals non-overlapping.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[1,2],[2,3]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> You don&#39;t need to remove any of the intervals since they&#39;re already non-overlapping.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 10<sup>5</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= start<sub>i</sub> &lt; end<sub>i</sub> &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        vector<pair<int,int>> inter;
        int n=intervals.size();
        for(int i=0;i<n;i++)
        {
            inter.push_back({intervals[i][0],intervals[i][1]});
        }
        sort(inter.begin(),inter.end());
        stack<pair<int,int>> st;
        st.push(inter[0]);
        for(int i=1;i<n;i++)
        {
            if(inter[i].first>=st.top().second)
            {
                st.push(inter[i]);
            }
            else
            {
                pair<int,int> temp=st.top();
                st.pop();
                temp.first=max(inter[i].first,temp.first);
                temp.second=min(temp.second,inter[i].second);
                st.push(temp);


            }
        }
        return n-st.size();
    }
};
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