<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>

bool cmp(vector<int> &a, vector<int> &b){
    return a[0] < b[0];
}
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) { 

        sort(intervals.begin(), intervals.end(), cmp);
        int ans = 0;
        int n = intervals.size();

        stack<vector<int> >s;
        
        for(int i = n - 1; i >= 0; i--) s.push(intervals[i]);

        while(s.size() > 1){
            auto a = s.top();
            s.pop();

            auto b = s.top();
            s.pop();

            if(a[1] > b[0]){ // if overlapping then remove that one whose end time is maximum bcs if we retain that one then chances of overlapping future increases 
                ans++;
                if(a[1] > b[1]) s.push(b);
                else s.push(a);
            }
            else s.push(b); // if not then next time we have to compare with second one so remove first
        }

        return ans;
    }
};
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